Question

Light of wavelength \(\lambda\) strikes a photoelectric surface and electrons are ejected with energy \(E\). If the wavelength is changed to \(\lambda'\), the energy becomes twice the original value. Which one of the following relations is true for \(\lambda\) and \(\lambda'\)?

• A. \( \lambda>\lambda'>\dfrac{\lambda}{2} \)
• B. \( \dfrac{\lambda}{4}<\lambda'<\dfrac{\lambda}{2} \)
• C. \( \lambda = 2\lambda' \)
• D. \( \dfrac{\lambda}{3}>\lambda'>\dfrac{\lambda}{4} \)

Hint:

In photoelectric effect, kinetic energy increases when wavelength decreases.

Answer 1

The Correct Option is A

Step 1: Photoelectric equation.
\[ E = h\nu - \phi = \frac{hc}{\lambda} - \phi \]
Step 2: Given condition for doubled energy.
\[ 2E = \frac{hc}{\lambda'} - \phi \]
Step 3: Eliminating work function \(\phi\).
Subtracting the two equations,
\[ E = hc\!\left(\frac{1}{\lambda'} - \frac{1}{\lambda}\right) \]
Step 4: Interpreting the inequality.
For energy to increase, wavelength must decrease, so \(\lambda'<\lambda\). But doubling of energy does not require halving of wavelength, hence \(\lambda'>\lambda/2\).
Step 5: Conclusion.
\[ \lambda>\lambda'>\frac{\lambda}{2} \]