Question

Threshold frequency for photoelectric effect from a metallic surface corresponds to a wavelength of 6000 \({Å}\). The photoelectric work function for the metal is \( h = 6.6 \times 10^{-34} \, {Js} \):

• A. \( 1.5 \times 10^{-19} \, {J} \)
• B. \( 2.7 \times 10^{-18} \, {J} \)
• C. \( 5.4 \times 10^{-18} \, {J} \)
• D. \( 4.5 \times 10^{-19} \, {J} \)
• E. \( 3.3 \times 10^{-19} \, {J} \)

Hint:

The work function \( W \) is related to the threshold frequency \( f_0 \) by \( W = h f_0 \). You can find \( f_0 \) using \( f_0 = \frac{c}{\lambda} \).

Answer 1

Answer: (E)

Step 1: The photoelectric work function \( W \) is related to the threshold frequency \( f_0 \) by: \[ W = h f_0 \] Step 2: The frequency \( f_0 \) can be calculated from the wavelength \( \lambda = 6000 \, {Å} = 6000 \times 10^{-10} \, {m} \) using the relation: \[ f_0 = \frac{c}{\lambda} \] where \( c = 3 \times 10^8 \, {m/s} \) is the speed of light. \[ f_0 = \frac{3 \times 10^8}{6000 \times 10^{-10}} = 5 \times 10^{13} \, {Hz} \] Step 3: Substitute \( f_0 = 5 \times 10^{13} \, {Hz} \) and \( h = 6.6 \times 10^{-34} \, {Js} \) into the equation for the work function: \[ W = 6.6 \times 10^{-34} \times 5 \times 10^{13} = 3.3 \times 10^{-19} \, {J} \] Thus, the photoelectric work function is \( 3.3 \times 10^{-19} \, {J} \).