Problem

A metal with a work function of 3.0 eV is illuminated by light of wavelength $3 \times 10^{-7} \, \mathrm{m}$. Calculate:
(a) The threshold frequency ($f_0$)
(b) The maximum kinetic energy ($KE_{\text{max}}$) of photoelectrons
(c) The stopping potential ($V_0$)

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Data

• Planck’s constant: $h = 6.626 \times 10^{-34} \, \mathrm{Js}$
• Speed of light: $c = 3 \times 10^8 \, \mathrm{ms^{-1}}$
• Wavelength: $\lambda = 3 \times 10^{-7} \, \mathrm{m}$
• Work function: $\phi = 3.0 \, \mathrm{eV} = 4.806 \times 10^{-19} \, \mathrm{J}$
• Charge of an electron: $e = 1.602 \times 10^{-19} \, \mathrm{C}$

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Prerequisite Concepts

1. Threshold frequency:

$$f_0 = \frac{\phi}{h}$$

2. Maximum kinetic energy:

$$KE_{\text{max}} = hf - \phi = \frac{hc}{\lambda} - \phi$$

3. Stopping potential:

$$V_0 = \frac{KE_{\text{max}}}{e}$$

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Solution

(a) Threshold Frequency ($f_0$)

Using the formula:

$$f_0 = \frac{\phi}{h}$$

Substitute values:

$$f_0 = \frac{4.806 \times 10^{-19} \, \mathrm{J}}{6.626 \times 10^{-34} \, \mathrm{Js}}$$ $$f_0 = 0.724 \times 10^{15} \, \mathrm{Hz}$$

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(b) Maximum Kinetic Energy ($KE_{\text{max}}$)

First, calculate the energy of the incident photon:

$$hf = \frac{hc}{\lambda}$$ $$hf = \frac{6.626 \times 10^{-34} \, \mathrm{Js} \times 3 \times 10^8 \, \mathrm{ms^{-1}}}{3 \times 10^{-7} \, \mathrm{m}}$$ $$hf = 6.626 \times 10^{-19} \, \mathrm{J}$$

Now, calculate $KE_{\text{max}}$:

$$KE_{\text{max}} = hf - \phi$$ $$KE_{\text{max}} = 6.626 \times 10^{-19} \, \mathrm{J} - 4.806 \times 10^{-19} \, \mathrm{J}$$ $$KE_{\text{max}} = 1.820 \times 10^{-19} \, \mathrm{J}$$

Convert to electron volts:

$$KE_{\text{max}} = \frac{1.820 \times 10^{-19}}{1.602 \times 10^{-19}} \, \mathrm{eV}$$ $$KE_{\text{max}} = 1.14 \, \mathrm{eV}$$

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(c) Stopping Potential ($V_0$)

Using the formula:

$$V_0 = \frac{KE_{\text{max}}}{e}$$

Substitute values:

$$V_0 = \frac{1.820 \times 10^{-19} \, \mathrm{J}}{1.602 \times 10^{-19} \, \mathrm{C}}$$ $$V_0 = 1.14 \, \mathrm{V}$$

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Answer

(a) Threshold frequency ($f_0$) = $0.724 \times 10^{15} \, \mathrm{Hz}$
(b) Maximum kinetic energy ($KE_{\text{max}}$) = $1.14 \, \mathrm{eV}$
(c) Stopping potential ($V_0$) = $1.14 \, \mathrm{V}$