Question:

An electron and a photon each have a wavelength of 1.00 nm. Find (a) their momenta, (b) the energy of the photon, and (c) the kinetic energy of electron.

Answer:

(a) Momentum of electron = h/λ = (6.626 x 10^-34 Js)/(1.00 x 10^-9 m) = 6.626 x 10^-25 kg m/s

Momentum of photon = h/λ = (6.626 x 10^-34 Js)/(1.00 x 10^-9 m) = 6.626 x 10^-25 kg m/s

(b) Energy of photon = hf = (6.626 x 10^-34 Js) x (3.00 x 10^8 m/s) / (1.00 x 10^-9 m) = 1.99 x 10^-19 J

(c) Kinetic energy of electron = (1/2) mv^2 = (1/2)(9.11 x 10^-31 kg)(3.00 x 10^8 m/s)^2 = 4.33 x 10^-14 J

Question:

The work function for a certain metal is 4.2 eV. Will this metal give photoelectric emission for incident radiation of wavelength 330 nm?

Answer:

1. Convert the wavelength 330 nm to energy: E = hc/λ = (6.626 x 10^-34 Js x 3 x 10^8 m/s) / (3.3 x 10^-7 m) = 6.1 x 10^-19 J

2. Compare the energy of the incident radiation to the work function of the metal: 6.1 x 10^-19 J < 4.2 eV = 6.6 x 10^-19 J

3. Conclusion: No, this metal will not give photoelectric emission for incident radiation of wavelength 330 nm.

Question:

Show that the wavelength of electromagnetic radiation is equal to the de Broglie wavelength of its quantum (photon).

Answer:

1. Recall that the wavelength of electromagnetic radiation is equal to the speed of light divided by the frequency of the wave.

λ = c/f

2. Also recall that the de Broglie wavelength of a quantum (photon) is equal to Planck’s constant divided by its momentum.

λ = h/p

3. Since the momentum of a photon is equal to its frequency multiplied by Planck’s constant, the de Broglie wavelength can be rewritten as:

λ = h/fh

4. Substituting this expression into the equation for the wavelength of electromagnetic radiation yields:

λ = c/fh

5. Since h is a constant, this equation can be simplified to:

λ = c/f

6. This is the same equation as the one for the wavelength of electromagnetic radiation, thus showing that the wavelength of electromagnetic radiation is equal to the de Broglie wavelength of its quantum (photon).

Question:

Monochromatic radiation of wavelength 640.2nm(1nm=109m) from a neon lamp irradiates photosensitive material made of caesium on tungsten. The stopping voltage is measured to be 0.54 V. The source is replaced by an iron source and its 427.2nm line irradiates the same photo-cell. Predict the new stopping voltage.

Answer:

1. Calculate the energy of the monochromatic radiation from the neon lamp using the equation E = hc/λ, where h is Planck’s constant (6.626 x 10-34 Js), c is the speed of light (3 x 108ms-1), and λ is the wavelength of the radiation (640.2nm).

2. Calculate the work function of the caesium on tungsten photo-sensitive material using the equation W = eV, where e is the charge of an electron (1.6 x 10-19C) and V is the stopping voltage (0.54V).

3. Calculate the energy of the monochromatic radiation from the iron source using the equation E = hc/λ, where h is Planck’s constant (6.626 x 10-34 Js), c is the speed of light (3 x 108ms-1), and λ is the wavelength of the radiation (427.2nm).

4. Calculate the new stopping voltage using the equation V = W/e, where W is the work function of the photo-sensitive material (calculated in step 2) and e is the charge of an electron (1.6 x 10-19C).

Question:

Find the (a) maximum frequency (b) Minimum wavelength of X-rays produced by 30 kV electrons.

Answer:

(a) Maximum frequency:

Step 1: Calculate the energy of the electrons.

E = 30 kV = 30,000 volts = 30,000 x 1.6 x 10^-19 J

Step 2: Calculate the frequency of the X-rays.

f = E/h

where h is Planck’s constant (6.63 x 10^-34 J s)

f = (30,000 x 1.6 x 10^-19 J)/(6.63 x 10^-34 J s)

f = 4.5 x 10^14 Hz

Therefore, the maximum frequency of the X-rays produced by 30 kV electrons is 4.5 x 10^14 Hz.

(b) Minimum wavelength:

Step 1: Calculate the frequency of the X-rays.

f = 4.5 x 10^14 Hz

Step 2: Calculate the wavelength of the X-rays.

λ = c/f

where c is the speed of light (3 x 10^8 m/s)

λ = (3 x 10^8 m/s)/(4.5 x 10^14 Hz)

λ = 6.67 x 10^-7 m

Therefore, the minimum wavelength of the X-rays produced by 30 kV electrons is 6.67 x 10^-7 m.

Question:

The wavelength of light from the spectral emission line of sodium is 589 nm. Find the kinetic energy at which (a) an electron, and (b) a neutron would have the same de Broglie wavelength.

Answer:

a) Kinetic energy of an electron = (h^2) / (2mλ^2) = (6.626 x 10^-34 Js)^2 / (2 x 9.109 x 10^-31 kg x (589 x 10^-9 m)^2) = 2.4 x 10^-19 J

b) Kinetic energy of a neutron = (h^2) / (2mλ^2) = (6.626 x 10^-34 Js)^2 / (2 x 1.675 x 10^-27 kg x (589 x 10^-9 m)^2) = 3.7 x 10^-13 J

Question:

A mercury lamp is a convenient source for studying frequency dependence of photoelectric emission, since, it gives a number of spectral lines ranging from the UV to the red end of the visible spectrum. In our experiment with rubidium photo-cell, the following lines from a mercury source were used λ1​=3650A˚,λ2​=4358A˚,λ3​=4358A˚,λ4​=5461A˚,λ5​=6907A˚.
The stopping voltages, respectively, were measured to be: V01​=1.28V, V02​=0.95V, V03​=0.74V, V04​=0.16V,V05​=0V,
Determine the value of Plancks constant h, the threshold frequency and work function for the material.

Answer:

1. Calculate the frequency of the spectral lines.

λ1​=3650A˚, f1​=8.4 x 10^14 Hz λ2​=4358A˚, f2​=6.9 x 10^14 Hz λ3​=4358A˚, f3​=6.9 x 10^14 Hz λ4​=5461A˚, f4​=5.5 x 10^14 Hz λ5​=6907A˚, f5​=4.3 x 10^14 Hz

2. Calculate the threshold frequency.

The threshold frequency is the frequency at which the stopping voltage is zero. In this case, the threshold frequency is f5​=4.3 x 10^14 Hz.

3. Calculate the work function.

The work function is the difference between the stopping voltage and the frequency of the spectral line.

V01​=1.28V, f1​=8.4 x 10^14 Hz, W1​=1.28V - 8.4 x 10^14 Hz = -8.4 x 10^14 eV V02​=0.95V, f2​=6.9 x 10^14 Hz, W2​=0.95V - 6.9 x 10^14 Hz = -6.9 x 10^14 eV V03​=0.74V, f3​=6.9 x 10^14 Hz, W3​=0.74V - 6.9 x 10^14 Hz = -6.9 x 10^14 eV V04​=0.16V, f4​=5.5 x 10^14 Hz, W4​=0.16V - 5.5 x 10^14 Hz = -5.5 x 10^14 eV V05​=0V, f5​=4.3 x 10^14 Hz, W5​=0V - 4.3 x 10^14 Hz = -4.3 x 10^14 eV

The average work function is -6.1 x 10^14 eV.

4. Calculate Plancks constant.

The value of Plancks constant can be calculated using the equation h=W/f, where W is the work function and f is the frequency.

h=W/f h=-6.1 x 10^14 eV / 4.3 x 10^14 Hz h=-1.4 x 10^-34 Js

Question:

The threshold frequency for a certain metal is 3.3×10^14Hz. If light of frequency 8.2×1014Hz is incident on the metal, predict the cutoff voltage for the photoelectric emission.

Answer:

Step 1: Calculate the difference between the threshold frequency (3.3×10^14Hz) and the frequency of the incident light (8.2×1014Hz).

Difference = 8.2×1014Hz - 3.3×10^14Hz = 4.9×1014Hz

Step 2: Use the formula hf = eV to calculate the cutoff voltage.

V = hf/e

V = (6.626×10^-34 Js × 4.9×1014Hz)/(1.602×10^-19 C)

V = 0.065V

Question:

What is the de Broglie wavelength of (a) a bullet of mass 0.040 kg travelling at the speed of 1.0 km/s, (b) a ball of mass 0.060 kg moving at a speed of 1.0 m/s, and (c) a dust particle of mass 1.0×10^9kg drifting with a speed of 2.2 m/s?

Answer:

(a) De Broglie wavelength = h/mv = (6.63×10^-34 Js)/(0.040 kg)(1000 m/s) = 1.65×10^-36 m

(b) De Broglie wavelength = h/mv = (6.63×10^-34 Js)/(0.060 kg)(1 m/s) = 1.10×10^-34 m

(c) De Broglie wavelength = h/mv = (6.63×10^-34 Js)/(1.0×10^-9 kg)(2.2 m/s) = 3.00×10^-26 m

Question:

What is the (a) momentum, (b) speed, and (c) de broglie wavelength of an electron with kinetic energy of 120 eV.

Answer:

(a) Momentum = sqrt(2mK) = sqrt(29.1110^-311201.610^-19) = 1.3910^-24 kg m/s

(b) Speed = sqrt(2K/m) = sqrt(21201.610^-19/9.1110^-31) = 1.82*10^6 m/s

(c) De Broglie Wavelength = h/p = 6.62610^-34/1.3910^-24 = 4.77*10^-11 m

Question:

Why is Prem determined not to return to his village?

Answer:

1. Prem may have experienced a traumatic event or had a bad experience in his village.
2. He may have been mistreated or discriminated against in his village.
3. He may have had a falling out with the people in his village.
4. He may have had a disagreement with the leaders in his village.
5. He may have had a difficult relationship with his family in his village.
6. He may have had a negative opinion of the lifestyle or culture in his village.
7. He may have wanted to pursue a different career or lifestyle away from his village.

Question:

Find the typical de Broglie wavelength associated with a He atom in helium gas at room temperature (27°C) and 1 atm pressure; and compare it with the mean separation between two atoms under these conditions.

Answer:

1. Calculate the temperature in Kelvin: 27°C = 300 K

2. Calculate the mass of a He atom: m = 4.0026 amu

3. Calculate the de Broglie wavelength: λ = h/(mv) where h = 6.626 x 10-34 J.s, m = 4.0026 amu, and v = the root-mean-square (rms) speed of the He atom at 300 K.

4. Calculate the rms speed of the He atom: v = (3RT/M)^1/2 where R = 8.314 J/mol.K and M = 4.0026 amu

5. Substitute the values into the de Broglie equation: λ = (6.626 x 10-34 J.s)/(4.0026 amu x (3 x 8.314 J/mol.K x 300 K/4.0026 amu)^1/2)

6. Calculate the de Broglie wavelength: λ = 2.26 x 10-10 m

7. Calculate the mean separation between two atoms at room temperature and 1 atm pressure: d = (3/4πn)^1/3 where n = the number of atoms/volume.

8. Calculate the number of atoms/volume: n = PV/RT where P = 1 atm, V = 22.4 L/mol, R = 8.314 J/mol.K, and T = 300 K.

9. Substitute the values into the mean separation equation: d = (3/4π x (1 atm x 22.4 L/mol)/(8.314 J/mol.K x 300 K))^1/3

10. Calculate the mean separation between two atoms: d = 3.68 x 10-10 m

11. Compare the de Broglie wavelength with the mean separation between two atoms: The de Broglie wavelength (2.26 x 10-10 m) is smaller than the mean separation between two atoms (3.68 x 10-10 m).

Question:

Light of intensity 10−5Wm2 falls on a sodium photo-cell of surface area 2 cm^2. Assuming that the top 5 layers of sodium absorb the incident energy, estimate time required for photoelectric emission in the wave-picture of radiation. The work function for the metal is given to be about 2 eV. What is the implication of your answer?

Answer:

1. Calculate the total energy absorbed by the sodium photo-cell: Total energy absorbed = 10^-5 Wm^2 x 2 cm^2 = 2 x 10^-5 J

2. Calculate the total energy required for photoelectric emission: Total energy required = Work Function x Charge of an electron = 2 eV x 1.6 x 10^-19 C = 3.2 x 10^-19 J

3. Calculate the time required for photoelectric emission: Time required = Total energy required/ Total energy absorbed = 3.2 x 10^-19 J/2 x 10^-5 J = 1.6 x 10^-14 s

4. Implication of the answer: The time required for photoelectric emission is very small, indicating that the emission of electrons from the sodium photo-cell is almost instantaneous when light of intensity 10^-5 Wm^2 is incident on it.

Question:

An electron gun with its collector at a potential of 100 V fires out electrons in a spherical bulb containing hydrogen gas at low pressure (~ 10−2 mm of Hg). A magnetic field of 2.83×10^4T curves the path of the electrons in a circular orbit of radius 12.0 cm. (The path can be viewed because the gas ions in the path focus the beam by attracting electrons, and emitting light by electron capture; this method is known as the fine beam tube method.) Determine e/m from the data.

Answer:

1. e/m is the ratio of the electron’s charge (e) to its mass (m).

2. The electron gun is used to fire electrons in a spherical bulb containing hydrogen gas at low pressure.

3. A magnetic field of 2.83×10^4T is used to curve the path of the electrons in a circular orbit of radius 12.0 cm.

4. The fine beam tube method is used to view the path of the electrons by focusing the beam and emitting light by electron capture.

5. To determine e/m from the data, we need to calculate the circumference of the circular orbit (C) and the period of the electron’s orbit (T).

6. The circumference of the circular orbit can be calculated using the formula C = 2πr, where r is the radius of the orbit (12 cm).

7. The period of the electron’s orbit can be calculated using the formula T = 2π√(L/B), where L is the length of the orbit (C) and B is the strength of the magnetic field (2.83×10^4T).

8. Finally, we can calculate e/m using the formula e/m = (C/T)^2.

Question:

a) Obtain the de Broglie wavelength of a neutron of kinetic energy 150 eV. As you have seen in Exercise 11.31, an electron beam of this energy is suitable for crystal diffraction experiments. Would a neutron beam of the same energy be equally suitable? Explain.(mn​=1.675×10^27kg) (b) Obtain the de Broglie wavelength associated with thermal neutrons at room temperature (27oC). Hence, explain why a fast neutron beam needs to be thermalised with the environment before it can be used for neutron diffraction experiments. Exercise 11.31: Crystal diffraction experiments can be performed using X-rays, or electrons accelerated through appropriate voltage. Which probe has greater energy? (For quantitative comparison, take the wave length of the probe equal to 1 A∘A∘ which is of the order of inter-atomic spacing in the lattice) (me​=9.11×10^−31kg)

Answer:

a) The de Broglie wavelength of a neutron of kinetic energy 150 eV is given by the equation λ=h/√2mek, where h is Planck’s constant, m is the mass of the neutron and k is the kinetic energy. Therefore, the de Broglie wavelength of the neutron is λ= (6.626 × 10^−34 J.s) / √(2 × 1.675 × 10^−27 kg × 150 eV) = 5.6 × 10^−12 m.

Yes, a neutron beam of the same energy would be equally suitable for crystal diffraction experiments. This is because the de Broglie wavelength of the neutron is of the same order as the inter-atomic spacing in the lattice, which is required for crystal diffraction experiments.

b) The de Broglie wavelength associated with thermal neutrons at room temperature (27oC) is given by the equation λ=h/√2mkT, where h is Planck’s constant, m is the mass of the neutron, k is Boltzmann’s constant and T is the temperature of the environment. Therefore, the de Broglie wavelength of the thermal neutron is λ= (6.626 × 10^−34 J.s) / √ (2 × 1.675 × 10^−27 kg × 1.38 × 10^−23 J/K × 273 K) = 0.31 × 10^−12 m.

Fast neutrons need to be thermalised with the environment before they can be used for neutron diffraction experiments because their de Broglie wavelength is much larger than the inter-atomic spacing in the lattice. Thermalising the neutrons reduces their energy and therefore their de Broglie wavelength, which makes them suitable for the experiment.

Question:

In an accelerator experiment on high-energy collisions of electrons with positrons, a certain event is interpreted as annihilation of an electron-positron pair of total energy 10.2 BeV into two γ-rays of equal energy. What is the wavelength associated with each γ-ray? (1BeV=109eV)

Answer:

Step 1: Convert 10.2 BeV to eV. Answer: 10.2 BeV = 1.02 x 1011 eV

Step 2: Calculate the energy of each gamma-ray. Answer: Each gamma-ray has an energy of 5.1 x 1010 eV.

Step 3: Use the formula E=hc/λ to calculate the wavelength of each gamma-ray. Answer: λ = hc/E = (6.626 x 10-34 Js)(3.00 x 108 m/s)/(5.1 x 1010 eV) = 5.2 x 10-12 m

Question:

The photoelectric cut-off voltage in a certain experiment is 1.5 V. What is the maximum kinetic energy of photoelectrons emitted?

Answer:

Answer: Step 1: Calculate the maximum kinetic energy of the photoelectrons using the equation KEmax = hf - W, where h is Planck’s constant (6.626 x 10-34 Js), f is the frequency of the radiation, and W is the work function of the material.

Step 2: Calculate the frequency of the radiation using the equation f = V/h, where V is the cut-off voltage (1.5 V).

Step 3: Calculate the work function of the material using the equation W = hf - KEmax, where h is Planck’s constant (6.626 x 10-34 Js) and KEmax is the maximum kinetic energy of the photoelectrons.

Step 4: Substitute the values for h, f, and W into the equation KEmax = hf - W and solve for KEmax.

The maximum kinetic energy of the photoelectrons is 1.2 x 10-19 J.

Question:

The work function of caesium metal is 2.14 eV. When the light of frequency 6×1014Hz is incident on the metal surface, photoemission of electrons occurs. Calculate: (a) the maximum kinetic energy of the emitted electrons, (b) stopping potential, (c) the maximum speed of the emitted photoelectrons?

Answer:

(a) The maximum kinetic energy of the emitted electrons is equal to the work function of caesium metal (2.14 eV).

(b) The stopping potential is equal to the energy of the incident photon (hf) minus the work function of the metal (2.14 eV).

Stopping potential = hf - 2.14 eV

Stopping potential = (6.626 x 10^-34 x 6 x 10^14) - (2.14 x 1.602 x 10^-19)

Stopping potential = 0.964 V

(c) The maximum speed of the emitted photoelectrons can be calculated using the equation vmax = √(2KE/m), where KE is the kinetic energy of the electrons and m is the mass of the electrons.

vmax = √(2 x 2.14 x 1.602 x 10^-19 / 9.11 x 10^-31)

vmax = 1.35 x 10^7 m/s

Question:

Calculate the (a) Momentum
(b) De Broglie wavelength of the electrons accelerated through a potential difference of 56 V.

Answer:

(a) Momentum:

Step 1: Calculate the kinetic energy of the electrons.

KE = 1/2 mv^2 = 1/2 * (9.11 x 10^-31 kg) * (V/L)^2 KE = 1/2 * (9.11 x 10^-31 kg) * (56 V/1 m)^2 KE = 3.17 x 10^-20 J

Step 2: Calculate the momentum of the electrons.

p = sqrt(2mKE) p = sqrt(2 * (9.11 x 10^-31 kg) * (3.17 x 10^-20 J)) p = 1.54 x 10^-24 kg m/s

(b) De Broglie Wavelength:

Step 1: Calculate the momentum of the electrons.

p = sqrt(2mKE) p = sqrt(2 * (9.11 x 10^-31 kg) * (3.17 x 10^-20 J)) p = 1.54 x 10^-24 kg m/s

Step 2: Calculate the de Broglie wavelength.

λ = h/p λ = (6.63 x 10^-34 J s) / (1.54 x 10^-24 kg m/s) λ = 4.30 x 10^-11 m

Question:

Light of frequency 7.21×10^14Hz is incident on a metal surface. Electrons with a maximum speed of 6.0×10^5m/s are ejected from the surface. What is the threshold frequency for photoemission of electrons?

Answer:

Step 1: Calculate the maximum kinetic energy of the ejected electrons using the equation E = 1/2 mv2

E = 1/2 (9.1 x 10-31 kg) (6.0 x 10^5 m/s)2

E = 2.7 x 10-19 J

Step 2: Calculate the threshold frequency using the equation hf = E

hf = 2.7 x 10-19 J

f = 4.02 x 10^14 Hz

Threshold frequency = 4.02 x 10^14 Hz

Question:

A 100W sodium lamp radiates energy uniformly in all directions. The lamp is located at the center of a large sphere that absorbs all the sodium light which is incident on it. The wavelength of the sodium light is 589 nm. (a) What is the energy per photon associated with the sodium light? (b) At what rate are the photons delivered to the sphere?

Answer:

(a) The energy per photon associated with the sodium light is calculated using the equation E = hc/λ, where h is Planck’s constant (6.626 x 10^-34 J*s), c is the speed of light (3 x 10^8 m/s), and λ is the wavelength of the sodium light (589 nm). Plugging in the values, we get E = 4.81 x 10^-19 J/photon.

(b) The rate at which the photons are delivered to the sphere is calculated using the equation P = EI, where P is the power (in watts), E is the energy per photon, and I is the intensity of the sodium light. Since the sodium lamp radiates energy uniformly in all directions, the intensity of the light is equal to the power output of the lamp divided by 4π, where π is the constant 3.14. Thus, the rate of photon delivery is P = (4.81 x 10^-19 J/photon)(100 W)/(4π) = 1.21 x 10^17 photons/s.

Question:

Answer the following questions: (a) Quarks inside protons and neutrons are thought to carry fractional charges [(+2/3)e;(1/3)e]. Why do they not show up in Millikan’s oil-drop experiment ? (b) What is so special about the combination e/m? Why do we not simply talk of e and m separately? (c) Why should gases be insulators at ordinary pressures and start conducting at very low pressures? (d) Every metal has a definite work function. Why do all photoelectrons not come out with the same energy if incident radiation is monochromatic? Why is there an energy distribution of photoelectrons? (e) The energy and momentum of an electron are related to the frequency and wavelength of the associated matter wave by the relations: E=hν, p = λh​. But while the value of λ is physically significant, the value of ν (and therefore, the value of the phase speed n λ) has no physical significance. Why?

Answer:

(a) Quarks inside protons and neutrons are thought to carry fractional charges [(+2/3)e;(1/3)e]. Why do they not show up in Millikan’s oil-drop experiment?

Quarks are too small to be detected in Millikan’s oil-drop experiment. The experiment relies on the use of charged particles that are large enough to be seen and measured.

(b) What is so special about the combination e/m? Why do we not simply talk of e and m separately?

The combination e/m is special because it is a constant, meaning the ratio of the charge of the particle to its mass is always the same. This allows us to measure the mass of a particle by measuring its charge. If we were to talk of e and m separately, then the ratio would not be constant and it would be difficult to measure the mass of a particle.

(c) Why should gases be insulators at ordinary pressures and start conducting at very low pressures?

At ordinary pressures, the molecules in a gas are spaced far enough apart that there are no free electrons available to carry electric current. When the pressure is reduced, the molecules become closer together, allowing for the creation of free electrons which can carry electric current, thus making the gas a conductor.

(d) Every metal has a definite work function. Why do all photoelectrons not come out with the same energy if incident radiation is monochromatic? Why is there an energy distribution of photoelectrons?

The work function of a metal is the minimum energy required to eject an electron from the metal’s surface. The incident radiation may have a higher energy, but the electrons can only absorb the energy up to the work function. Any energy above the work function is not absorbed and is instead emitted as kinetic energy, resulting in an energy distribution of photoelectrons.

(e) The energy and momentum of an electron are related to the frequency and wavelength of the associated matter wave by the relations: E=hν, p = λh​. But while the value of λ is physically significant, the value of ν (and therefore, the value of the phase speed n λ) has no physical significance. Why?

The phase speed (nλ) is the speed of the wave, which is not physically significant because the wave is not moving. The wavelength (λ) is physically significant because it is related to the momentum of the particle (p = λh).

Question:

The work function for the following metals is given: Na: 2.75 eV; K: 2.30 eV; Mo: 4.17 eV; Ni: 5.15 eV. Which of these metals will not give photoelectric emission for a radiation of wavelength 3300 from a He-Cd laser placed 1 m away from the photocell? What happens if the laser is brought nearer and placed 50 cm away ?

Answer:

1. The work function of the given metals are: Na: 2.75 eV; K: 2.30 eV; Mo: 4.17 eV; Ni: 5.15 eV.

2. The energy of the radiation from the He-Cd laser is given by: E = hc/λ = (6.626 x 10-34 x 3 x 108)/(3300 x 10-10) = 1.99 eV.

3. Since the energy of the radiation is lower than the work function of Na (2.75 eV), K (2.30 eV) and Mo (4.17 eV), these metals will not give photoelectric emission for the given radiation.

4. However, the energy of the radiation is higher than the work function of Ni (5.15 eV). Hence, Ni will give photoelectric emission for the given radiation.

5. If the laser is brought nearer and placed 50 cm away, the energy of the radiation will increase. This will result in all the metals giving photoelectric emission.

Question:

What is the de Broglie wavelength of a nitrogen molecule in air at 300 K? Assume that the molecule is moving with the root-mean square speed of molecules at this temperature. (Atomic mass of nitrogen = 14.0076 u)

Answer:

Step 1: Calculate the root-mean square speed of molecules at 300 K.

Step 2: Calculate the momentum of the nitrogen molecule in air at 300 K.

Step 3: Calculate the de Broglie wavelength of the nitrogen molecule using the equation λ = h/p, where h is Planck’s constant and p is the momentum.

Question:

(a) An X-ray tube produces a continuous spectrum of radiation with its short wavelength end at 0.45 . What is the maximum energy of a photon in the radiation? (b) From your answer to (a), guess what order of accelerating voltage(for electrons) is required in such a tube?

Answer:

a) The maximum energy of a photon in the radiation is equal to hc/λ, where h is Planck’s constant, c is the speed of light, and λ is the wavelength. Therefore, the maximum energy of a photon in the radiation is equal to hc/0.45, or approximately 4.4 x 10^-19 Joules.

b) The accelerating voltage required in such a tube would be equal to the maximum energy of the photon, or approximately 4.4 x 10^-19 Joules.

Question:

Ultraviolet light of wavelength 2271Ao from a 100W mercury source irradiates a photo-cell made of molybdenum metal. If the stopping potential is −1.3V, estimate the work function of the metal. How would the photo-cell respond to a high-intensity(∼105Wm - 2) red light of wavelength 6328A0 produced by a He-Ne laser? (h=6.63×10^−34 js,c=3×10^8m/s)

Answer:

1. Calculate the energy of the ultraviolet light: E=hc/λ = (6.63×10^−34 Js)(3×10^8m/s)/2271Ao = 2.86×10^-19 J

2. Calculate the number of photons emitted by the mercury source: n=100W/E = (100W)/(2.86×10^-19 J) = 3.49×10^20 photons

3. Calculate the work function of the molybdenum metal: W=eV = (1.6×10^-19 C)(-1.3V) = -2.08×10^-19 J

4. The photo-cell would not respond to the high-intensity red light of wavelength 6328Ao produced by the He-Ne laser because the energy of the red light is less than the work function of the molybdenum metal. E=hc/λ = (6.63×10^−34 Js)(3×10^8m/s)/6328Ao = 1.05×10^-19 J, which is less than the work function of -2.08×10^-19 J.

Question:

(a) For what kinetic energy of a neutron will the associated de Broglie wavelength be 1.40×10^−10m? (b) Find the de Broglie wavelength of a neutron in thermal equilibrium with matter having an average kinetic energy of (3/2)kT at 300K.

Answer:

(a) The kinetic energy of a neutron associated with a de Broglie wavelength of 1.40×10^−10m is calculated using the equation E = hv, where E is the kinetic energy, h is Planck’s constant (6.626 x 10^-34 Js) and v is the frequency of the neutron.

v = c/λ

where c is the speed of light (3 x 10^8 m/s) and λ is the de Broglie wavelength (1.40 x 10^-10 m).

Substituting these values into the equation, we get:

E = hv = (6.626 x 10^-34 Js)(3 x 10^8 m/s)/(1.40 x 10^-10 m) = 2.84 x 10^-14 J

(b) The de Broglie wavelength of a neutron in thermal equilibrium with matter having an average kinetic energy of (3/2)kT at 300K is calculated using the equation E = (3/2)kT, where E is the kinetic energy, k is Boltzmann’s constant (1.38 x 10^-23 J/K) and T is the temperature (300K).

Substituting these values into the equation, we get:

E = (3/2)kT = (3/2)(1.38 x 10^-23 J/K)(300K) = 2.07 x 10^-21 J

The de Broglie wavelength of a neutron is calculated using the equation λ = h/p, where h is Planck’s constant (6.626 x 10^-34 Js) and p is the momentum of the neutron.

The momentum of the neutron is calculated using the equation p = mv, where m is the mass of the neutron (1.67 x 10^-27 kg) and v is the velocity of the neutron.

The velocity of the neutron is calculated using the equation v = √2E/m, where E is the kinetic energy (2.07 x 10^-21 J) and m is the mass of the neutron (1.67 x 10^-27 kg).

Substituting these values into the equation, we get:

v = √2E/m = √2(2.07 x 10^-21 J)/(1.67 x 10^-27 kg) = 3.38 x 10^5 m/s

Substituting this value into the equation for momentum, we get:

p = mv = (1.67 x 10^-27 kg)(3.38 x 10^5 m/s) = 5.65 x 10^-22 kg m/s

Substituting this value into the equation for de Broglie wavelength, we get:

λ = h/p = (6.626 x 10^-34 Js)/(5.65 x 10^-22 kg m/s) = 1.17 x 10^-12 m

Question:

Why did Makara dislike tortoises, snakes and lizards? Write a line about each.

Answer:

Makara disliked tortoises because they moved too slowly for her liking. Makara disliked snakes because they were too slimy and creepy for her. Makara disliked lizards because they were too scaly and strange for her tastes.

Question:

Light of wavelength 488 nm is produced by an argon laser which is used in the photoelectric effect. When light from this spectral line is incident on the emitter, the stopping (cut-off) potential of photoelectrons is 0.38 V. Find the work function(in eV) of the material from which the emitter is made.

Answer:

1. Convert the wavelength of light from nanometers to meters: λ = 488 nm = 4.88 x 10^-7 m

2. Calculate the frequency of the light: f = c/λ where c is the speed of light (3.00 x 10^8 m/s) f = 3.00 x 10^8 m/s / 4.88 x 10^-7 m f = 6.12 x 10^14 Hz

3. Calculate the energy of the light: E = hf where h is Planck’s constant (6.626 x 10^-34 J s) E = 6.626 x 10^-34 J s x 6.12 x 10^14 Hz E = 4.04 x 10^-19 J

4. Convert the energy of the light from joules to electron volts: 1 J = 6.24 x 10^18 eV E = 4.04 x 10^-19 J x 6.24 x 10^18 eV/J E = 2.51 x 10^-1 eV

5. Calculate the work function: Φ = E - V where E is the energy of the light (2.51 x 10^-1 eV) and V is the stopping potential (0.38 V) Φ = 2.51 x 10^-1 eV - 0.38 V Φ = 2.13 x 10^-1 eV

Therefore, the work function of the material from which the emitter is made is 2.13 x 10^-1 eV.

Question:

In an experiment on photoelectric effect, the slope of the cut-off voltage versus frequency of incident light is found to be 4.12×10^−15Vs. Calculate the value of Plancks constant.

Answer:

Step 1: The equation for the photoelectric effect is given by:

E = hf - φ

Where E is the cut-off voltage, h is Plancks constant, f is the frequency of incident light, and φ is the work function of the material.

Step 2: Rearrange the equation to solve for h:

h = E/f + φ

Step 3: Substitute the given values for E and f:

h = (4.12 x 10^-15 Vs)/f + φ

Step 4: Solve for Plancks constant:

h = 4.12 x 10^-15 Vs/f + φ

h = 4.12 x 10^-15 Vs/f + φ

h = 4.12 x 10^-15 Vs/f

h = 4.12 x 10^-15 Vs/f

h = 6.626 x 10^-34 Js

Question:

Estimating the following two numbers should be interesting. The first number will tell you why radio engineers do not need to worry much about photons. The second number tells you why our eye can never count photons, even in barely detectable light. (a) The number of photons emitted per second by a medium wave transmitter of 10 kW power, emitting radio waves of wave length 500 m. (b) The number of photons entering the pupil of our eye per second corresponding to the minimum intensity of white light that we humans can perceive ( 1010Wm^2). Take the area of the pupil to be about 0.4cm^2, and the average frequency of white light to be about 6×10^14Hz.

Answer:

a) First, we need to calculate the energy per photon. This can be done by using the formula E = hf, where E is the energy per photon, h is Planck’s constant (6.63 x 10^-34 Js) and f is the frequency of the wave. In this case, the frequency is 500 m, which is equal to 5 x 10^8 Hz. Therefore, the energy per photon is 3.315 x 10^-26 J.

Next, we need to calculate the number of photons emitted per second. This can be done by using the formula N = P/E, where N is the number of photons emitted per second, P is the power of the transmitter (10 kW) and E is the energy per photon. Therefore, the number of photons emitted per second is 3 x 10^37.

b) First, we need to calculate the intensity of the light. This can be done by using the formula I = P/A, where I is the intensity of the light, P is the power of the light (1010 Wm^2) and A is the area of the pupil (0.4 cm^2). Therefore, the intensity of the light is 2.5 x 10^12 Wm^-2.

Next, we need to calculate the energy per photon. This can be done by using the formula E = hf, where E is the energy per photon, h is Planck’s constant (6.63 x 10^-34 Js) and f is the frequency of the wave. In this case, the frequency is 6 x 10^14 Hz. Therefore, the energy per photon is 4.038 x 10^-19 J.

Finally, we need to calculate the number of photons entering the pupil of our eye per second. This can be done by using the formula N = I/E, where N is the number of photons entering the pupil per second, I is the intensity of the light and E is the energy per photon. Therefore, the number of photons entering the pupil per second is 6.2 x 10^31.

Question:

An electron microscope uses electrons accelerated by a voltage of 50 kV. Determine the de-Broglie wavelength associated with the electrons. Taking other factors, such as numerical aperture etc. to be same, how does the resolving power of an electron microscope compare with that of an optical microscope which uses yellow light? (λy​=5.9×10^−7m)

Answer:

1. Calculate the de-Broglie wavelength associated with the electrons:

de-Broglie wavelength (λ) = h/mv

where h is Planck’s constant (6.62607×10^−34 Js), m is the mass of an electron (9.10938356×10^−31 kg) and v is the velocity of the electrons (which can be calculated using the given voltage of 50 kV):

v = √2E/m

where E is the energy of the electrons (which can be calculated using the given voltage of 50 kV):

E = Vq

where V is the voltage (50 kV) and q is the charge of an electron (1.60217662×10^−19 C).

Therefore, the de-Broglie wavelength can be calculated as follows:

λ = h/mv

= (6.62607×10^−34 Js)/(9.10938356×10^−31 kg)√(2(50 kV)(1.60217662×10^−19 C))

= 1.22×10^−10 m

2. The resolving power of an electron microscope is much higher than that of an optical microscope which uses yellow light. This is because the wavelength of the electrons is much shorter than the wavelength of yellow light (1.22×10^−10 m compared to 5.9×10^−7 m).

Question:

The energy flux of sunlight reaching the surface of the earth is 1.388×10^3W/m2. How many photons (nearly) per square metre are incident on the Earth per second? Assume that the photons in the sunlight have an average wavelength of 550 nm.

Answer:

1. Convert 1.388×10^3 W/m2 to Joules per second per square metre: 1.388×10^3 W/m2 = 1.388×10^3 J/s/m2

2. Calculate the energy of one photon with a wavelength of 550 nm: E = hc/λ E = (6.626×10^-34 Js x 3.00×10^8 m/s)/(550×10^-9 m) E = 3.95×10^-19 J

3. Calculate the number of photons per square metre per second: Number of photons = 1.388×10^3 J/s/m2/3.95×10^-19 J Number of photons = 3.51×10^21 photons/m2/s