Question

The stopping potential when a metal surface is illuminated by light of wavelength \( \lambda \) is \( 15\,V \). The stopping potential when the same surface is illuminated by light of wavelength \( 4\lambda \) is \( 3\,V \). The ratio of threshold wavelength to the initial incident wavelength \( \lambda \) is:

• A. \( 1:16 \)
• B. \( 1:15 \)
• C. \( 16:1 \)
• D. \( 15:1 \)

Hint:

In photoelectric effect, stopping potential is directly related to frequencyAlways use two equations to eliminate constants like \( hc/e \).

Answer 1

The Correct Option is C

Step 1: Use photoelectric equation.
\[ eV = \frac{hc}{\lambda} - \frac{hc}{\lambda_0} \]
where \( \lambda_0 \) is threshold wavelength.

Step 2: Write equation for first case.
\[ 15 = \frac{hc}{e}\left(\frac{1}{\lambda} - \frac{1}{\lambda_0}\right) \]

Step 3: Write equation for second case.
\[ 3 = \frac{hc}{e}\left(\frac{1}{4\lambda} - \frac{1}{\lambda_0}\right) \]

Step 4: Subtract second equation from first.
\[ 15 - 3 = \frac{hc}{e}\left(\frac{1}{\lambda} - \frac{1}{4\lambda}\right) \]
\[ 12 = \frac{hc}{e}\left(\frac{3}{4\lambda}\right) \]

Step 5: Solve for \( \frac{hc}{e\lambda} \).
\[ \frac{hc}{e\lambda} = 16 \]

Step 6: Substitute into first equation.
\[ 15 = 16 - \frac{hc}{e\lambda_0} \]
\[ \frac{hc}{e\lambda_0} = 1 \]

Step 7: Find ratio.
\[ \frac{\lambda_0}{\lambda} = \frac{16}{1} \]
\[ \boxed{16:1} \]