Question

Photoelectrons are emitted from two similar metal plates when wavelengths \(λ_1\) and \(λ_2\) are incident on them (\(λ_1 = 1.5λ_2\)). If maximum kinetic energy of emitted photoelectrons is \(E_1\) and \(E_2\) respectively then

• A. \(E_1 < 2E_2/3\)
• B. \(E_1 = 2E_2/3\)
• C. \(E_1 = E_2/3\)
• D. \(E_1 = 2E_2\)

Hint:

Write Einstein equation for each wavelength and compare with the work function.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Einstein's equation: \(E_k=\dfrac{hc}{\lambda}-\phi\), where \(\phi>0\) is the work function. Both plates are similar, so \(\phi\) is the same.

Step 2: Write for each wavelength:
\(E_1=\dfrac{hc}{\lambda_1}-\phi=\dfrac{hc}{1.5\lambda_2}-\phi=\dfrac23\cdot\dfrac{hc}{\lambda_2}-\phi\). \(E_2=\dfrac{hc}{\lambda_2}-\phi\).

Step 3: Compare E1 with 2E2/3:
\(\dfrac23E_2=\dfrac23\cdot\dfrac{hc}{\lambda_2}-\dfrac23\phi\). So \(E_1-\dfrac23E_2=-\phi+\dfrac23\phi=-\dfrac\phi3<0\).

Step 4: Conclusion:
\(E_1<\dfrac{2E_2}{3}\). Option A.

Step 5: Why the other options are wrong.
B, C and D would need \(\phi=0\), which is not possible for a metal with a real threshold. They would hold only if the kinetic energy were proportional to the photon energy.

Final Answer:
E1 is less than 2 E2 / 3. \[ \boxed{\text{(A) }E_1<\dfrac{2E_2}{3}} \]