Question

The kinetic energy of electrons emitted, when radiation of frequency \( 1.0 \times 10^{15} \) Hz hits a metal, is \( 2 \times 10^{-19} \) J. What is the threshold frequency of the metal (in Hz)? (\( h = 6.6 \times 10^{-34} \) Js)

• A. \( 3.5 \times 10^{15} \)
• B. \( 3.3 \times 10^{14} \)
• C. \( 6.97 \times 10^{15} \)
• D. \( 6.97 \times 10^{14} \)

Hint:

Use the photoelectric equation: \( h f = h f_0 + K_{\text{max}} \).
- To find \( f_0 \), rearrange: \( f_0 = \frac{h f - K_{\text{max}}}{h} \).

Answer 1

The Correct Option is D

The photoelectric equation is given by: \[ h f = h f_0 + K_{\text{max}} \] where: - \( h \) is Planck’s constant (\( 6.6 \times 10^{-34} \) Js), - \( f \) is the incident frequency (\( 1.0 \times 10^{15} \) Hz), - \( f_0 \) is the threshold frequency, - \( K_{\text{max}} \) is the maximum kinetic energy of emitted electrons (\( 2 \times 10^{-19} \) J). 1. Rearrange the equation for \( f_0 \): \[ f_0 = \frac{h f - K_{\text{max}}}{h} \] 2. Substituting the values: \[ f_0 = \frac{(6.6 \times 10^{-34} \times 1.0 \times 10^{15}) - (2 \times 10^{-19})}{6.6 \times 10^{-34}} \] \[ = \frac{6.6 \times 10^{-19} - 2 \times 10^{-19}}{6.6 \times 10^{-34}} \] \[ = \frac{4.6 \times 10^{-19}}{6.6 \times 10^{-34}} \] \[ = 6.97 \times 10^{14} \text{ Hz} \] Thus, the correct answer is \(\boxed{6.97 \times 10^{14} \text{ Hz}}\).