Question

Light of wavelength \(\lambda\) strikes a photoelectric surface and electrons are ejected with an energy \(E\). If \(E\) is to be increased to exactly twice its original value, the wavelength changes to \(\lambda _{1}\) where

• A. \(\lambda _{1}\lt \frac{\lambda }{2}\)
• B. \(\lambda _{1}>\lambda\)
• C. \(\lambda _{1}>\frac{\lambda }{2}\lt \lambda\)
• D. \(\lambda _{1}=\frac{\lambda }{2}\)

Hint:

Use Einstein's photoelectric equation for both wavelengths and compare the photon energies.

Answer 1

The Correct Option is C

Step 1: Reason with Energies
Photon energy must increase, but by less than double, because the work function stays the same.

Step 2: Check the Limits
Doubling the photon energy would make $E$ larger than $2E$, so $\lambda_1$ cannot be $\lambda/2$ or less.

Step 3: Conclusion
Photon energy rises, so $\lambda_1<\lambda$, and rises by less than double, so $\lambda_1>\lambda/2$.

Final Answer:
The wavelength is shorter than the original but longer than half of it. \[ \boxed{\text{(C) }\lambda_{1}>\dfrac{\lambda}{2}\ \text{and}\ \lambda_{1}<\lambda} \]