Numerical Problems

1) Work Function and Maximum Wavelength $${\lambda }_{max}=\frac{hc}{\varphi }$$ Where,

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

Given:

• Work function, $$\varphi =4.2eV=(4.2eV)(1.6\times {10}^{-19}J/eV)=6.72\times {10}^{-19}J$$

Calculations $${\lambda }_{max}=\frac{(6.626\times {10}^{-34}Js)(3\times {10}^{8}m/s)}{6.72\times {10}^{-19}J}=\boxed{{\displaystyle 292nm}}$$

2) Work Function and Stopping Potential $$e{V}_s=hf-\varphi$$ Where,

• $$e{V}_s$$ is the stopping potential energy in joules (J).
• $$h$$ is Planck’s constant ($$6.626\times {10}^{-34}Js$$).
• $$f$$ is the frequency of the incident light in hertz (Hz).
• $$\varphi$$ is the work function in joules (J).

Given:

• Wavelength of light, $$\lambda =400nm=400\times {10}^{-9}m$$
• Stopping potential, $$V_s=0.5V=0.5J/C$$

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

Now, calculate the work function: $$e{V}_s=hf-\varphi$$ $$\varphi =hf-e{V}_s=(6.626\times {10}^{-34}Js)(7.5\times {10}^{14}Hz)-(0.5J)$$ $$\varphi =4.97\times {10}^{-19}J-0.5J=-0.000498J=\boxed{{\displaystyle (}}-0.31eV)$$

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 $$e{V}_{s1}=h{f}_1-\varphi$$ $$e{V}_{s2}=h{f}_2-\varphi$$

Dividing the second equation by the first equation:

$$\frac{e{V}_{s2}}{e{V}_{s1}}=\frac{h{f}_2-\varphi }{h{f}_1-\varphi }$$

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

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 $$K_{max}=hf-\varphi$$

Where,

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

Given:

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

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

Therefore, the maximum kinetic energy of emitted photoelectrons is $$2.63\times {10}^{-19}J$$

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 ($$6.626\times {10}^{-34}Js$$).
• $$f$$ is the frequency of incident light in hertz (Hz).

Given:

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

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

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

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

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

Given:

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

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