Answer

Potential of the electrons, $V=30 \mathrm{kV}=3 \times 10^{4} \mathrm{~V}$

Hence, energy of the electrons, $E=3 \times 10^{4} \mathrm{eV}$

Where,

$e=$ Charge on an electron $=1.6 \times 10^{-19} \mathrm{C}$

(a)Maximum frequency produced by the $\mathrm{X}$-rays $=v$

The energy of the electrons is given by the relation:

$E=h v$

Where,

$h=$ Planck’s constant $=6.626 \times 10^{-34} \mathrm{Js}$

$\therefore v=\frac{E}{h}$

$$ =\frac{1.6 \times 10^{-19} \times 3 \times 10^{4}}{6.626 \times 10^{-34}}=7.24 \times 10^{18} \mathrm{~Hz} $$

Hence, the maximum frequency of X-rays produced is $7.24 \times 10^{18} \mathrm{~Hz}$.

(b)The minimum wavelength produced by the X-rays is given as:

$$ \begin{aligned} & \lambda=\frac{c}{v} \\ & =\frac{3 \times 10^{8}}{7.24 \times 10^{18}}=4.14 \times 10^{-11} \mathrm{~m}=0.0414 \mathrm{~nm} \end{aligned} $$

Hence, the minimum wavelength of X-rays produced is $0.0414 \mathrm{~nm}$.

Answer

Work function of caesium metal, $\phi_{0}=2.14 \mathrm{eV}$

Frequency of light, $v=6.0 \times 10^{14} \mathrm{~Hz}$

(a)The maximum kinetic energy is given by the photoelectric effect as:

$$ K=h v-\phi_{0} $$

Where,

$$ \begin{aligned} & h=\text { Planck’s constant }=6.626 \times 10^{-34} \mathrm{Js} \\ & \therefore K=\frac{6.626 \times 10^{34} \times 6 \times 10^{14}}{1.6 \times 10^{-19}}-2.14 \\ & \quad=2.485-2.140=0.345 \mathrm{eV} \end{aligned} $$

Hence, the maximum kinetic energy of the emitted electrons is $0.345 \mathrm{eV}$.

(b)For stopping potential $V_{0}$, we can write the equation for kinetic energy as:

$$ \begin{aligned} & K=e V_{0} \\ & \therefore V_{0}=\frac{K}{e} \\ & \quad=\frac{0.345 \times 1.6 \times 10^{-19}}{1.6 \times 10^{-19}}=0.345 \mathrm{~V} \end{aligned} $$

Hence, the stopping potential of the material is $0.345 \mathrm{~V}$.

(c)Maximum speed of the emitted photoelectrons $=v$

Hence, the relation for kinetic energy can be written as:

$$ K=\frac{1}{2} m v^{2} $$

Where,

$$ \begin{aligned} m & =\text { Mass of an electron }=9.1 \times 10^{-31} \mathrm{~kg} \\ v^{2} & =\frac{2 K}{m} \\ & =\frac{2 \times 0.345 \times 1.6 \times 10^{-19}}{9.1 \times 10^{-31}}=0.1104 \times 10^{12} \\ \therefore v & =3.323 \times 10^{5} \mathrm{~m} / \mathrm{s}=332.3 \mathrm{~km} / \mathrm{s} \end{aligned} $$

Hence, the maximum speed of the emitted photoelectrons is $332.3 \mathrm{~km} / \mathrm{s}$.

Answer

Photoelectric cut-off voltage, $V_{0}=1.5 \mathrm{~V}$

The maximum kinetic energy of the emitted photoelectrons is given as:

$$ K_{e}=e V_{0} $$

Where,

$e=$ Charge on an electron $=1.6 \times 10^{-19} \mathrm{C}$

$$ \begin{aligned} \therefore K_{e} & =1.6 \times 10^{-19} \times 1.5 \\ & =2.4 \times 10^{-19} \mathrm{~J} \end{aligned} $$

Therefore, the maximum kinetic energy of the photoelectrons emitted in the given experiment is $2.4 \times 10^{-19} \mathrm{~J}$.

Answer

Wavelength of the monochromatic light, $\lambda=632.8 \mathrm{~nm}=632.8 \times 10^{-9} \mathrm{~m}$

Power emitted by the laser, $P=9.42 \mathrm{~mW}=9.42 \times 10^{-3} \mathrm{~W}$

Planck’s constant, $h=6.626 \times 10^{-34} \mathrm{Js}$

Speed of light, $c=3 \times 10^{8} \mathrm{~m} / \mathrm{s}$

Mass of a hydrogen atom, $m=1.66 \times 10^{-27} \mathrm{~kg}$

(a)The energy of each photon is given as:

$$ \begin{aligned} E & =\frac{h c}{\lambda} \\ & =\frac{6.626 \times 10^{-34} \times 3 \times 10^{8}}{632.8 \times 10^{-9}}=3.141 \times 10^{-19} \mathrm{~J} \end{aligned} $$

The momentum of each photon is given as:

$$ \begin{aligned} P & =\frac{h}{\lambda} \\ & =\frac{6.626 \times 10^{-34}}{632.8}=1.047 \times 10^{-27} \mathrm{~kg} \mathrm{~m} \mathrm{~s}^{-1} \end{aligned} $$

(b)Number of photons arriving per second, at a target irradiated by the beam $=n$

Assume that the beam has a uniform cross-section that is less than the target area.

Hence, the equation for power can be written as:

$$ \begin{aligned} P & =n E \\ \therefore n & =\frac{P}{E} \\ & =\frac{9.42 \times 10^{-3}}{3.141 \times 10^{-19}} \approx 3 \times 10^{16} \text { photon } / \mathrm{s} \end{aligned} $$

(c)Momentum of the hydrogen atom is the same as the momentum of the photon, $p=1.047 \times 10^{-27} \mathrm{~kg} \mathrm{~m} \mathrm{~s}^{-1}$

Momentum is given as:

$p=m v$

Where,

$v=$ Speed of the hydrogen atom

$$ \begin{aligned} \therefore v & =\frac{p}{m} \\ & =\frac{1.047 \times 10^{-27}}{1.66 \times 10^{-27}}=0.621 \mathrm{~m} / \mathrm{s} \end{aligned} $$

Answer

The slope of the cut-off voltage $(V)$ versus frequency $(v)$ of an incident light is given as:

$\frac{V}{v}=4.12 \times 10^{-15} \mathrm{Vs}$

$V$ is related to frequency by the equation:

$h v=e V$

Where,

$e=$ Charge on an electron $=1.6 \times 10^{-19} \mathrm{C}$

$h=$ Planck’s constant

$\therefore h=e \times \frac{V}{v}$

$=1.6 \times 10^{-19} \times 4.12 \times 10^{-15}=6.592 \times 10^{-34} \mathrm{JS}$

Therefore, the value of Planck’s constant is $6.592 \times 10^{-34} \mathrm{Js}$.

Answer

Threshold frequency of the metal, $v_{0}=3.3 \times 10^{14} \mathrm{~Hz}$

Frequency of light incident on the metal, $v=8.2 \times 10^{14} \mathrm{~Hz}$

Charge on an electron, $e=1.6 \times 10^{-19} \mathrm{C}$

Planck’s constant, $h=6.626 \times 10^{-34} \mathrm{Js}$

Cut-off voltage for the photoelectric emission from the metal $=V_{0}$

The equation for the cut-off energy is given as:

$$ \begin{aligned} e V_{0} & =h\left(v-v_{0}\right) \\ V_{0} & =\frac{h\left(v-v_{0}\right)}{e} \\ & =\frac{6.626 \times 10^{-34} \times\left(8.2 \times 10^{14}-3.3 \times 10^{14}\right)}{1.6 \times 10^{-19}}=2.0292 \mathrm{~V} \end{aligned} $$

Therefore, the cut-off voltage for the photoelectric emission is $2.0292 \mathrm{~V}$.

Answer

Work function of the metal, $\phi_{0}=4.2 \mathrm{eV}$

Charge on an electron, $e=1.6 \times 10^{-19} \mathrm{C}$

Planck’s constant, $h=6.626 \times 10^{-34} \mathrm{Js}$

Wavelength of the incident radiation, $\lambda=330 \mathrm{~nm}=330 \times 10^{-9} \mathrm{~m}$

Speed of light, $c=3 \times 10^{8} \mathrm{~m} / \mathrm{s}$

The energy of the incident photon is given as:

$$ \begin{aligned} E & =\frac{h c}{\lambda} \\ & =\frac{6.626 \times 10^{-34} \times 3 \times 10^{8}}{330 \times 10^{-9}}=6.0 \times 10^{-19} \mathrm{~J} \\ & =\frac{6.0 \times 10^{-19}}{1.6 \times 10^{-19}}=3.76 \mathrm{eV} \end{aligned} $$

It can be observed that the energy of the incident radiation is less than the work function of the metal. Hence, no photoelectric emission will take place.

Answer

Frequency of the incident photon, $v=488 \mathrm{~nm}=488 \times 10^{-9} \mathrm{~m}$

Maximum speed of the electrons, $v=6.0 \times 10^{5} \mathrm{~m} / \mathrm{s}$

Planck’s constant, $h=6.626 \times 10^{-34} \mathrm{Js}$

Mass of an electron, $m=9.1 \times 10^{-31} \mathrm{~kg}$

For threshold frequency $v_{0}$, the relation for kinetic energy is written as:

$$ \begin{aligned} & \frac{1}{2} m v^{2}=h\left(v-v_{0}\right) \\ & v_{0}=v-\frac{m v^{2}}{2 h} \\ & \quad=7.21 \times 10^{14}-\frac{\left(9.1 \times 10^{-31}\right) \times\left(6 \times 10^{5}\right)^{2}}{2 \times\left(6.626 \times 10^{-34}\right)} \\ & \quad=7.21 \times 10^{14}-2.472 \times 10^{14} \\ & \quad=4.738 \times 10^{14} \mathrm{~Hz} \end{aligned} $$

Therefore, the threshold frequency for the photoemission of electrons is $4.738 \times 10^{14} \mathrm{~Hz}$.

Answer

Wavelength of light produced by the argon laser, $\lambda=488 \mathrm{~nm}$ $=488 \times 10^{-9} \mathrm{~m}$

Stopping potential of the photoelectrons, $V_{0}=0.38 \mathrm{~V}$

$$ \begin{aligned} & 1 \mathrm{eV}=1.6 \times 10^{-19} \mathrm{~J} \\ & \therefore V_{0}=\frac{0.38}{1.6 \times 10^{-19}} \mathrm{eV} \end{aligned} $$

Planck’s constant, $h=6.6 \times 10^{-34} \mathrm{Js}$

Charge on an electron, $e=1.6 \times 10^{-19} \mathrm{C}$

Speed of light, $c=3 \times 10 \mathrm{~m} / \mathrm{s}$

From Einstein’s photoelectric effect, we have the relation involving the work function $\Phi_{0}$ of the material of the emitter as:

$$ \begin{aligned} & e V_{0}=\frac{h c}{\lambda}-\phi_{0} \\ & \phi_{0}=\frac{h c}{\lambda}-e V_{0} \\ & \quad=\frac{6.6 \times 10^{-34} \times 3 \times 10^{8}}{1.6 \times 10^{-19} \times 488 \times 10^{-9}}-\frac{1.6 \times 10^{-19} \times 0.38}{1.6 \times 10^{-19}} \\ & \quad=2.54-0.38=2.16 \mathrm{eV} \end{aligned} $$

Therefore, the material with which the emitter is made has the work function of $2.16 \mathrm{eV}$.

Answer

(a)Mass of the bullet, $m=0.040 \mathrm{~kg}$

Speed of the bullet, $v=1.0 \mathrm{~km} / \mathrm{s}=1000 \mathrm{~m} / \mathrm{s}$

Planck’s constant, $h=6.6 \times 10^{-34} \mathrm{Js}$

De Broglie wavelength of the bullet is given by the relation:

$$ \begin{aligned} & \lambda=\frac{h}{m v} \\ & =\frac{6.6 \times 10^{-34}}{0.040 \times 1000}=1.65 \times 10^{-35} \mathrm{~m} \end{aligned} $$

Mass of the ball, $m=0.060 \mathrm{~kg}$

Speed of the ball, $v=1.0 \mathrm{~m} / \mathrm{s}$

De Broglie wavelength of the ball is given by the relation:

$$ \begin{aligned} & \lambda=\frac{h}{m v} \\ & =\frac{6.6 \times 10^{-34}}{0.060 \times 1}=1.1 \times 10^{-32} \mathrm{~m} \end{aligned} $$

(c)Mass of the dust particle, $m=1 \times 10^{-9} \mathrm{~kg}$

Speed of the dust particle, $v=2.2 \mathrm{~m} / \mathrm{s}$

De Broglie wavelength of the dust particle is given by the relation:

$$ \lambda=\frac{h}{m v} $$

$=\frac{6.6 \times 10^{-34}}{2.2 \times 1 \times 10^{-9}}=3.0 \times 10^{-25} \mathrm{~m}$

Answer

The momentum of a photon having energy $(h v)$ is given as:

$$ \begin{align*} & p=\frac{h v}{c}=\frac{h}{\lambda} \\ & \lambda=\frac{h}{p} \tag{i} \end{align*} $$

Where,

$\lambda=$ Wavelength of the electromagnetic radiation

$c=$ Speed of light

$h=$ Planck’s constant

De Broglie wavelength of the photon is given as:

$\lambda=\frac{h}{m v}$

But $p=m v$

$\therefore \lambda=\frac{h}{p}$

Where,

$m=$ Mass of the photon

$v=$ Velocity of the photon

Hence, it can be inferred from equations (i) and (ii) that the wavelength of the electromagnetic radiation is equal to the de Broglie wavelength of the photon.