Question

Photocathode work function is \( 1 \, \text{eV} \). Light of wavelength \( \lambda = 3000 \, \text{Å} \) falls on it. The photoelectron comes out with a maximum velocity of \( 1 \times 10^6 \, \text{m/s} \). What is the energy of the photon?

• A. 1.0 eV
• B. 1.5 eV
• C. 2.0 eV
• D. 3.0 eV

Hint:

To find the energy of the photon, use the work function and the kinetic energy of the emitted photoelectron.

Answer 1

The Correct Option is C

Step 1: Energy of the photon.
The energy of the photon can be calculated using the equation: \[ E = \phi + K.E. \] where \( \phi \) is the work function and \( K.E. \) is the kinetic energy of the emitted electron.

Step 2: Calculate kinetic energy.
The kinetic energy is given by: \[ K.E. = \frac{1}{2} m v^2 \] where \( m = 9.1 \times 10^{-31} \, \text{kg} \) and \( v = 1 \times 10^6 \, \text{m/s} \). Substituting these values, we get: \[ K.E. = 4.55 \times 10^{-19} \, \text{J} = 2.8 \, \text{eV}. \] Thus, the total energy of the photon is: \[ E = 1 \, \text{eV} + 2 \, \text{eV} = 3 \, \text{eV}. \]