Question

Two light waves of wavelengths 600 nm and 200 nm are incident on a metal surface. The maximum velocity of photoelectrons produced due to one wavelength is \(\frac{1}{3}\) of the maximum velocity of the photoelectrons produced due to the other wavelength. Find the work function of the metal.

• A. \(\frac{hc}{8} \times 10^7 \, \text{J}\)
• B. \(\frac{8}{hc} \times 10^7 \, \text{J}\)
• C. \(\frac{hc}{4} \times 10^7 \, \text{J}\)
• D. \(\frac{hc}{9} \times 10^7 \, \text{J}\)

Hint:

For photoelectric effect with velocity ratio: relate \(v_{\text{max}}^2 \propto K_{\text{max}} = h\nu - \phi\) to solve for \(\phi\).

Answer 1

The Correct Option is D

Step 1: Turn the velocity ratio into a kinetic energy ratio.
Since $K=\tfrac12mv^2$, if $v_{600} = \tfrac13 v_{200}$ then $K_{600} = \tfrac19 K_{200}$.
Step 2: Write both photoelectric equations.
\[ K_{600} = \frac{hc}{600}-\phi, \qquad K_{200} = \frac{hc}{200}-\phi \]
Step 3: Substitute $K_{200}=9K_{600}$ and eliminate $K_{600}$ directly.
\[ 9\left(\frac{hc}{600}-\phi\right) = \frac{hc}{200}-\phi \implies \frac{9hc}{600}-\frac{hc}{200} = 8\phi \]
Step 4: Combine the wavelength terms over one denominator, then convert nm to m.
\[ \frac{9}{600}-\frac{1}{200} = 0.015-0.005 = 0.01\ \text{nm}^{-1} = 1\times10^7\ \text{m}^{-1} \]
\[ \phi = \frac{hc\times10^7}{8} = \frac{hc}{8}\times10^7\ \text{J} \]
\[ \boxed{\frac{hc}{8}\times10^7\ \text{J}} \]