Numerical Problems

1) Work Function and Maximum Wavelength $$λ_{max}=\frac{hc}{\phi}$$ Where,

• $$λ_{max}$$ is the maximum wavelength of light in meters (m).
• $$h$$ is Planck’s constant ($$6.626 × 10^{−34}Js$$).
• $$c$$ is the speed of light ($$3 × 10^8 m/s$$).
• $$\phi$$ is the work function in joules (J).

Given:

• Work function, $$\phi = 4.2 eV = (4.2 eV)(1.6 × 10^{−19}J/eV) = 6.72 × 10^{−19} J$$

Calculations $$λ_{max}=\frac{(6.626 × 10^{−34}Js)(3 × 10^8m/s)}{6.72 × 10^{−19}J}=\boxed{292 nm}$$

2) Work Function and Stopping Potential $$eV_s=hf-\phi$$ Where,

• $$eV_s$$ is the stopping potential energy in joules (J).
• $$h$$ is Planck’s constant ($$6.626 × 10^{−34}Js$$).
• $$f$$ is the frequency of the incident light in hertz (Hz).
• $$\phi$$ is the work function in joules (J).

Given:

• Wavelength of light, $$λ = 400 nm = 400 × 10^{−9}m$$
• Stopping potential, $$V_s = 0.5 V = 0.5 J/C$$

Calculations First, calculate the frequency of light using the formula: $$f=\frac{c}{λ}=\frac{3 × 10^8 m/s}{400 × 10^{−9}m}=7.5 × 10^{14} Hz$$

Now, calculate the work function: $$eV_s=hf-\phi$$ $$\phi=hf−eV_s=(6.626×10^{−34}Js)(7.5×10^{14}Hz)−(0.5J)$$ $$\phi=4.97×10^{−19}J−0.5J=−0.000498J=\boxed(−0.31 eV)$$

Therefore, the work function of the metal is 0.31 eV (negative value indicates that the given wavelength is below the threshold wavelength).

3) Effect of Frequency on Stopping Potential $$eV_{s1}=hf_1−\phi$$ $$eV_{s2}=hf_2−\phi$$

Dividing the second equation by the first equation:

$$\frac{eV_{s2}}{eV_{s1}}=\frac{hf_2−\phi}{hf_1−\phi}$$

Since $$\phi$$ is the same for both frequencies:

$$\frac{V_{s2}}{V_{s1}}=\frac{f_2}{f_1}$$ Given:

• Stopping potential for frequency $$f_1, V_{s1} = 2V$$
• The frequency of light is doubled, $$f_2 = 2f_1$$

Calculations: $$\frac{V_{s2}}{2V}=\frac{2f_1}{f_1}=2$$ $$V_{s2}=2(2V)=\boxed{4V}$$

Therefore, the stopping potential will become 4V when the frequency of light is doubled.

4) Stopping Potential and Light Intensity $$V_s\propto I$$

Where,

• $$V_s$$ is the stopping potential in volts (V).
• $$I$$ is the intensity of light.

Given:

• Intensity of light from source 1, $$I_1$$
• Intensity of light from source 2, $$I_2 = 2I_1$$
• Stopping potential for source 1, $$V_{s1}$$

Calculations As the intensity of light increases, the stopping potential also increases. Since the intensity of light from source 2 is twice the intensity of light from source 1:

$$\frac{V_{s2}}{V_{s1}}=\frac{I_2}{I_1}=\frac{2I_1}{I_1}=2$$

Therefore, the ratio of the stopping potentials is 2, indicating that the stopping potential for source 2 is twice the stopping potential for source 1.

5) Maximum Kinetic Energy of Emitted Photoelectrons $$K_{max}=hf−\phi$$

Where,

• $$K_{max}$$ is the maximum kinetic energy of emitted electrons in joules (J).
• $$h$$ is Planck’s constant ($$6.626 × 10^{−34}Js$$).
• $$f$$ is the frequency of incident light in hertz (Hz).
• $$\phi$$ is the work function in joules (J).

Given:

• Wavelength of light, $$\lambda = 300 nm = 300 × 10^{−9}m$$
• Work function, $$\phi = 2.5 eV = (2.5 eV)(1.6 × 10^{−19}J/eV) = 4.0 × 10^{−19} J$$

Calculations First, calculate the frequency of light using the formula: $$f=\frac{c}{λ}=\frac{3 × 10^8 m/s}{300 × 10^{−9}m}=10^{15} Hz$$ Now, calculate the maximum kinetic energy: $$K_{max}=hf−\phi=(6.626×10^{−34}Js)(10^{15}Hz)−4.0×10^{−19}J=\boxed{2.63×10^{−19}J}$$

Therefore, the maximum kinetic energy of emitted photoelectrons is $$2.63 × 10^{−19}J$$

6) Number of Photons $$n=\frac{I}{hf}$$ Where,

• $$n$$ is the number of photons per second.
• $$I$$ is the intensity of light in watts (W).
• $$h$$ is Planck’s constant ($$6.626 × 10^{−34}Js$$).
• $$f$$ is the frequency of incident light in hertz (Hz).

Given:

• Intensity of light, $$I=1\mu A=10^{−6}A$$
• Wavelength of light, $$\lambda = 550 nm = 550 × 10^{−9}m$$
• Work function, $$\phi = 1.8 eV = (1.8 eV)(1.6 × 10^{−19}J/eV) = 2.88 × 10^{−19} J$$

Calculations First, calculate the frequency of light using the formula: $$f=\frac{c}{λ}=\frac{3 × 10^8 m/s}{550 × 10^{−9}m}=5.45 × 10^{14} Hz$$

Now, calculate the number of photons per second: $$n=\frac{I}{hf}=\frac{10^{−6}W}{(6.626 × 10^{−34}Js)(5.45 × 10^{14} Hz)}$$ $$n=\boxed{2.94 × 10^{13}\text{ photons/s}}$$

7) De Broglie Wavelength of Emitted Electrons $$\lambda=\frac{h}{p}=\frac{h}{\sqrt{2m_ek_{max}}}$$ Where,

• $$\lambda$$ is the de Broglie wavelength of the emitted electrons in meters (m).
• $$h$$ is Planck’s constant ($$6.626 × 10^{−34}Js$$).
• $$m_e$$ is the mass of an electron ($$9.109 × 10^{−31} kg$$).
• $$K_{max}$$ is the maximum kinetic energy of emitted electrons in joules (J).

Given:

• Wavelength of incident light, $$\lambda = 250 nm = 250 × 10^{−9}m$$
• Work function, $$\phi = 4.5 eV = (4.5 eV)(1.6 × 10^{−19}J/eV) = 7.2 × 10^{−19} J$$

Calculations First, calculate the maximum kinetic energy of emitted electrons using the formula: $$K_{max}=hf−\phi= \frac{hc}{\lambda}-\phi