Question

When a photosensitive metal surface is illuminated with radiation of wavelength ' $\lambda_1$ ', the stopping potential is ' $V_1$ '. If the same surface is illuminated with radiation of wavelength ' $3\lambda_1$ ', the stopping potential is $\frac{V_1}{6}$. The threshold wavelength for the photosensitive metal surface is

• A. $\frac{3}{2} \lambda_1$
• B. $2\lambda_1$
• C. $5\lambda_1$
• D. $6\lambda_1$

Hint:

For photoelectric effect: \[ eV_s=hc\left(\frac{1}{\lambda}-\frac{1}{\lambda_0}\right) \] If two stopping potentials are given, convert the relation into a simple equation using reciprocals.

Answer 1

The Correct Option is B

Concept:
Einstein's photoelectric equation in terms of stopping potential is: \[ eV_s=\frac{hc}{\lambda}-\phi \] If threshold wavelength is \(\lambda_0\), then: \[ \phi=\frac{hc}{\lambda_0} \] So, \[ eV_s=hc\left(\frac{1}{\lambda}-\frac{1}{\lambda_0}\right) \] ip

Step 1: Write equation for wavelength \(\lambda_1\).
\[ eV_1=hc\left(\frac{1}{\lambda_1}-\frac{1}{\lambda_0}\right) \] ip

Step 2: Write equation for wavelength \(3\lambda_1\).
Given stopping potential becomes: \[ \frac{V_1}{6} \] So, \[ e\frac{V_1}{6}=hc\left(\frac{1}{3\lambda_1}-\frac{1}{\lambda_0}\right) \] ip

Step 3: Use ratio form.
Let \[ A=\frac{1}{\lambda_1},\qquad B=\frac{1}{\lambda_0} \] Then: \[ V_1 \propto A-B \] and \[ \frac{V_1}{6}\propto \frac{A}{3}-B \] So, \[ \frac{\frac{A}{3}-B}{A-B}=\frac{1}{6} \] ip

Step 4: Solve for \(B\).
\[ 6\left(\frac{A}{3}-B\right)=A-B \] \[ 2A-6B=A-B \] \[ A=5B \] \[ B=\frac{A}{5} \] Thus, \[ \frac{1}{\lambda_0}=\frac{1}{5\lambda_1} \] \[ \lambda_0=5\lambda_1 \] So the direct physics result is: \[ \boxed{5\lambda_1} \] which matches option (C), not option (B). ip The direct calculation gives:
\[ \boxed{5\lambda_1} \]