Question

Light of wavelength \(\lambda\) strikes a photo-sensitive surface and electrons are ejected with kinetic energy \(E\). If the kinetic energy is to be increased to \(2E\), the wavelength must be changed to \(\lambda'\) where

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

Hint:

In photoelectric effect, kinetic energy increases if wavelength decreases. But doubling energy does not always mean halving wavelength because of work function \(\phi\).

Answer 1

The Correct Option is C

Step 1: Use Einstein's photoelectric equation.
\[ K_{max} = h\nu - \phi = \frac{hc}{\lambda} - \phi \] Step 2: Write given condition.
Initially,
\[ E = \frac{hc}{\lambda} - \phi \] Finally,
\[ 2E = \frac{hc}{\lambda'} - \phi \] Step 3: Compare both equations.
\[ 2\left(\frac{hc}{\lambda} - \phi\right) = \frac{hc}{\lambda'} - \phi \] \[ \Rightarrow \frac{2hc}{\lambda} - 2\phi = \frac{hc}{\lambda'} - \phi \Rightarrow \frac{hc}{\lambda'} = \frac{2hc}{\lambda} - \phi \] Step 4: Conclude range of \(\lambda'\).
Since \(\phi>0\),
\[ \frac{hc}{\lambda'}<\frac{2hc}{\lambda} \Rightarrow \lambda'>\frac{\lambda}{2} \] Also to increase energy, \(\lambda'\) must decrease compared to \(\lambda\):
\[ \lambda'<\lambda \] Thus,
\[ \frac{\lambda}{2}<\lambda'<\lambda \] Final Answer: \[ \boxed{\dfrac{\lambda}{2}<\lambda'<\lambda} \]