Question

The maximum wavelength of light which causes photoelectric emission from a photosensitive metal surface is \( \lambda_0 \). Two light beams of wavelengths \( \frac{\lambda_0}{3} \) and \( \frac{\lambda_0}{9} \) incident on the metal surface. The ratio of the maximum velocities of the emitted photoelectrons is:

• A. \( 3:4 \)
• B. \( 1:3 \)
• C. \( 1:2 \)
• D. \( 2:3 \)

Hint:

For photoelectric emission, the kinetic energy is given by \( K.E. = \frac{hc}{\lambda} - \frac{hc}{\lambda_0} \), and velocity is proportional to the square root of kinetic energy.

Answer 1

The Correct Option is C

Step 1: Apply Energy and Photoelectric Equation The kinetic energy of the emitted photoelectron is given by: \[ K.E. = \frac{hc}{\lambda} - \frac{hc}{\lambda_0} \] Since \( K.E. \propto v^2 \), the velocity is given by: \[ v = \sqrt{\frac{2K.E.}{m}} \] Step 2: Compute Kinetic Energy For \( \lambda = \frac{\lambda_0}{3} \): \[ K.E._1 = \frac{hc}{\frac{\lambda_0}{3}} - \frac{hc}{\lambda_0} \] \[ K.E._1 = \frac{3hc}{\lambda_0} - \frac{hc}{\lambda_0} = \frac{2hc}{\lambda_0} \] For \( \lambda = \frac{\lambda_0}{9} \): \[ K.E._2 = \frac{hc}{\frac{\lambda_0}{9}} - \frac{hc}{\lambda_0} \] \[ K.E._2 = \frac{9hc}{\lambda_0} - \frac{hc}{\lambda_0} = \frac{8hc}{\lambda_0} \] Step 3: Compute Velocity Ratio \[ \frac{v_1}{v_2} = \sqrt{\frac{K.E._1}{K.E._2}} = \sqrt{\frac{2hc/\lambda_0}{8hc/\lambda_0}} \] \[ = \sqrt{\frac{1}{4}} = \frac{1}{2} \] Thus, the correct answer is \( 1:2 \).

Answer 2

Given:
- Threshold wavelength: \( \lambda_0 \)
- Two incident light wavelengths: \( \frac{\lambda_0}{3} \) and \( \frac{\lambda_0}{9} \)

We need to find the ratio of the maximum velocities of the photoelectrons emitted by these two wavelengths.

Step 1: Photoelectric equation:
\[ KE_{max} = h\nu - \phi \] where
- \( KE_{max} \) is the maximum kinetic energy of emitted electrons,
- \( h \) is Planck's constant,
- \( \nu \) is the frequency of incident light,
- \( \phi \) is the work function of the metal.

Step 2: Relation between wavelength and frequency:
\[ \nu = \frac{c}{\lambda} \] where \( c \) is the speed of light.

Step 3: Work function and threshold wavelength relation:
\[ \phi = h \nu_0 = \frac{hc}{\lambda_0} \]

Step 4: Calculate maximum kinetic energies for the two wavelengths:
For \( \lambda_1 = \frac{\lambda_0}{3} \):
\[ KE_1 = h \frac{c}{\lambda_1} - \frac{hc}{\lambda_0} = hc \left( \frac{1}{\lambda_1} - \frac{1}{\lambda_0} \right) \] \[ = hc \left( \frac{1}{\lambda_0/3} - \frac{1}{\lambda_0} \right) = hc \left( \frac{3}{\lambda_0} - \frac{1}{\lambda_0} \right) = \frac{2hc}{\lambda_0} \]

For \( \lambda_2 = \frac{\lambda_0}{9} \):
\[ KE_2 = h \frac{c}{\lambda_2} - \frac{hc}{\lambda_0} = hc \left( \frac{1}{\lambda_2} - \frac{1}{\lambda_0} \right) \] \[ = hc \left( \frac{1}{\lambda_0/9} - \frac{1}{\lambda_0} \right) = hc \left( \frac{9}{\lambda_0} - \frac{1}{\lambda_0} \right) = \frac{8hc}{\lambda_0} \]

Step 5: Using kinetic energy relation to velocity:
\[ KE = \frac{1}{2} m v^2 \Rightarrow v = \sqrt{\frac{2 KE}{m}} \]

Step 6: Ratio of velocities:
\[ \frac{v_1}{v_2} = \sqrt{\frac{KE_1}{KE_2}} = \sqrt{\frac{2hc/\lambda_0}{8hc/\lambda_0}} = \sqrt{\frac{2}{8}} = \sqrt{\frac{1}{4}} = \frac{1}{2} \]

Therefore, the ratio of maximum velocities of the photoelectrons is:
\[ v_1 : v_2 = 1 : 2 \]