Question

On a metal surface if light of wavelength $\lambda$ falls stopping potential for emitted photoelectron is $3V_0$ and if light of wavelength $2\lambda$ falls stopping potential is $V_0$. Find threshold wavelength :-

Answer 1

On a metal surface, if light of wavelength \( \lambda \) falls, the stopping potential for the emitted photoelectron is \( 3V_0 \). If light of wavelength \( 2\lambda \) falls, the stopping potential is \( V_0 \). Find the threshold wavelength.

Step 1: Using Einstein's photoelectric equation
Einstein's photoelectric equation relates the stopping potential \( V \), the energy of the incident photons, and the work function \( \phi \) of the metal: \[ E_{\text{photon}} = \phi + eV. \] The energy of the incident photon is related to its wavelength \( \lambda \) by: \[ E_{\text{photon}} = \frac{hc}{\lambda}, \] where \( h \) is Planck's constant and \( c \) is the speed of light.

Step 2: Apply for light of wavelength \( \lambda \)
For the first case where the stopping potential is \( 3V_0 \), the equation becomes: \[ \frac{hc}{\lambda} = \phi + e(3V_0). \] Hence, we have: \[ \frac{hc}{\lambda} = \phi + 3eV_0. \quad \text{(Equation 1)} \]

Step 3: Apply for light of wavelength \( 2\lambda \)
For the second case where the stopping potential is \( V_0 \), the equation becomes: \[ \frac{hc}{2\lambda} = \phi + eV_0. \] Hence, we have: \[ \frac{hc}{2\lambda} = \phi + eV_0. \quad \text{(Equation 2)} \]

Step 4: Solve the system of equations
Subtract Equation 2 from Equation 1: \[ \left( \frac{hc}{\lambda} - \frac{hc}{2\lambda} \right) = (3eV_0) - eV_0, \] \[ \frac{hc}{2\lambda} = 2eV_0. \] Thus, solving for \( \lambda \): \[ \frac{hc}{2\lambda} = 2eV_0 \implies \lambda = \frac{hc}{4eV_0}. \]

Final Answer: The threshold wavelength \( \lambda_{\text{threshold}} = 4\lambda \).