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: Understanding the Concept
Einstein's equation gives \(E=\dfrac{hc}{\lambda}-\phi\), where \(\phi\) is the work function.

Step 2: Write Both Cases
Initially \(\dfrac{hc}{\lambda}=E+\phi\).
For the new wavelength \(\dfrac{hc}{\lambda_1}=2E+\phi\).

Step 3: Compare with Half the Wavelength
For wavelength \(\lambda/2\) the photon energy would be \(2(E+\phi)=2E+2\phi\), which is larger than \(2E+\phi\).
So \(\lambda_1>\dfrac{\lambda}{2}\).

Step 4: Compare with the Original Wavelength
\(2E+\phi>E+\phi\), so the new photon energy is larger and \(\lambda_1<\lambda\).
Hence \(\dfrac{\lambda}{2}<\lambda_1<\lambda\), option (C).

Final Answer:
The new wavelength lies between half the original and the original wavelength. \[ \boxed{\text{(C) }\lambda_{1}>\dfrac{\lambda}{2}\ \text{and}\ \lambda_{1}<\lambda} \]