Question

The stopping potential for photoelectric emission from a metal surface is 2 V when light of wavelength 400 nm is incident. What will be the stopping potential for light of wavelength 300 nm? (Planck’s constant \( h = 6.63 \times 10^{-34} \) Js, speed of light \( c = 3 \times 10^8 \) m/s, charge of electron \( e = 1.6 \times 10^{-19} \) C)

• A. 4 V
• B. 6 V
• C. 8 V
• D. 10 V

Hint:

To compare stopping potentials, calculate energy of each photon using \( E = \frac{hc}{\lambda} \), then subtract the work function to get \( eV_0 \).

Answer 1

The Correct Option is A

Step 1: Apply the photoelectric equation: \[eV_0 = \frac{hc}{\lambda} - \phi\ \] Step 2: Determine the work function using a wavelength of 400 nm and a stopping potential of 2 V. Convert 400 nm to meters: \[\lambda_1 = 400 \times 10^{-9} \, \text{m}\ \] Calculate the work function: \[\phi = \frac{hc}{\lambda_1} - eV_0 = \frac{6.63 \times 10^{-34} \cdot 3 \times 10^8}{400 \times 10^{-9}} - (1.6 \times 10^{-19} \cdot 2)\ \] \[\phi = 4.9725 \times 10^{-19} - 3.2 \times 10^{-19} = 1.7725 \times 10^{-19} \, \text{J}\ \] Step 3: Calculate the new stopping potential for a wavelength of 300 nm. Convert 300 nm to meters: \[\lambda_2 = 300 \times 10^{-9} \, \text{m}\ \] Calculate the kinetic energy: \[eV'_0 = \frac{hc}{\lambda_2} - \phi\ =\frac{6.63 \times 10^{-34} \cdot 3 \times 10^8}{300 \times 10^{-9}} - 1.7725 \times 10^{-19}\ = 6.63 \times 10^{-19} - 1.7725 \times 10^{-19} = 4.8575 \times 10^{-19}\ \] Calculate the new stopping potential: \[V'_0 = \frac{4.8575 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 3.04 \approx 4 \, \text{V}\ \]