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

Step 1: Use Einstein's photoelectric equation: $eV_s = \frac{hc}{\lambda} - \phi$.
Where $V_s$ is stopping potential and $\phi$ is the work function ($\phi = \frac{hc}{\lambda_{th}}$).

Step 2: Write equations for both cases.
Case 1: $3eV_0 = \frac{hc}{\lambda} - \phi$ ---(1)
Case 2: $eV_0 = \frac{hc}{2\lambda} - \phi$ ---(2)

Step 3: Solve for $\phi$.
Multiply equation (2) by 3: $3eV_0 = \frac{3hc}{2\lambda} - 3\phi$.
Equate to equation (1): $\frac{hc}{\lambda} - \phi = \frac{3hc}{2\lambda} - 3\phi$.
$2\phi = \frac{3hc}{2\lambda} - \frac{hc}{\lambda} = \frac{3hc - 2hc}{2\lambda} = \frac{hc}{2\lambda}$.
$\phi = \frac{hc}{4\lambda}$.

Step 4: Relate $\phi$ to threshold wavelength $\lambda_{th}$.
$\frac{hc}{\lambda_{th}} = \frac{hc}{4\lambda} \implies \lambda_{th} = 4\lambda$.

Final Answer: Threshold wavelength is $4\lambda$.