Question

Light of wavelength \(\lambda\) strikes a photoelectric surface and electrons are ejected with energy \(E\). If \(E\) is to be increased to twice the original value, the wavelength changes to \(\lambda_1\)

• A. \(\lambda_1<\lambda/2\)
• B. \(\lambda_1 = \lambda\)
• C. \(\lambda_1>\lambda/2\)
• D. \(\lambda_1 = \lambda/2\)

Hint:

Due to work function, wavelength is always less than half for doubling energy.

Answer 1

The Correct Option is A

Concept:
Photoelectric equation: \[ E = \frac{hc}{\lambda} - \phi \] Step 1: Initial condition. \[ E = \frac{hc}{\lambda} - \phi \]
Step 2: Final condition. \[ 2E = \frac{hc}{\lambda_1} - \phi \]
Step 3: Compare equations. \[ 2\left(\frac{hc}{\lambda} - \phi\right) = \frac{hc}{\lambda_1} - \phi \] \[ \frac{2hc}{\lambda} - 2\phi = \frac{hc}{\lambda_1} - \phi \] \[ \frac{hc}{\lambda_1} = \frac{2hc}{\lambda} - \phi \]
Step 4: Conclusion. Since work function exists: \[ \lambda_1<\frac{\lambda}{2} \]