Question

A photoelectric surface is illuminated successively by monochromatic light of wavelength \(λ\) and \((λ/3)\). If the maximum kinetic energy of the emitted photo electrons in the second case is \(4\) times that in the first case, the work function of the surface of the material is (\(h\) = Planck's constant, \(c\) = speed of light)

• A. \(\frac{3hc}{λ}\)
• B. \(\frac{hc}{3λ}\)
• C. \(\frac{hc}{2λ}\)
• D. \(\frac{hc}{λ}\)

Hint:

Apply Einstein's equation twice and use the factor of 4 between the kinetic energies.

Answer 1

The Correct Option is B

Step 1: Einstein's Equation:
\(K=\dfrac{hc}\lambda-\phi\).

Step 2: Two Cases:
First case: \(K_1=\dfrac{hc}\lambda-\phi\). Second case, wavelength \(\dfrac\lambda3\): \(K_2=\dfrac{3hc}\lambda-\phi\).

Step 3: Use K2 = 4 K1:
\[ \frac{3hc}\lambda-\phi=4\left(\frac{hc}\lambda-\phi\right)\Rightarrow3\phi=\frac{hc}\lambda\Rightarrow\phi=\frac{hc}{3\lambda} \]

Step 4: Check the Options:
For \(\phi=\dfrac{hc}{3\lambda}\): \(K_1=\dfrac{2hc}{3\lambda}\) and \(K_2=\dfrac{8hc}{3\lambda}\), a ratio of 4. This is option (B). Other options give a different ratio.

Final Answer:
The work function is \(\dfrac{hc}{3\lambda}\), option (B). \[ \boxed{\text{(B) } \frac{hc}{3\lambda}} \]