Q. The energy that should be added to an electron, to reduce its de-Broglie wavelength from $10^{- 10}m$ to $0.5 \times 10^{- 10}m,$ will be
A) Four times the initial energy
B) Thrice the initial energy
C) Equal to the initial energy
D) Twice the initial energy
Explanation
(b) $\lambda = \frac{h}{\sqrt{2mE}} \Rightarrow \lambda \propto \frac{1}{\sqrt{E}} \Rightarrow \frac{\lambda_{1}}{\lambda_{2}} = \sqrt{\frac{E_{2}}{E_{1}}}$
$\Rightarrow \frac{10^{- 10}}{0.5 \times 10^{- 10}} = \sqrt{\frac{E_{2}}{E_{1}}} \Rightarrow E_{2} = 4E_{1}$
Hence added energy $= E_{2} - E_{1} = 3E_{1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.