Q. The energy that should be added to an electron to reduce its de-Broglie wavelength from 1 nm to 0.5 nm is
A) Four times the initial energy
B) Equal to the initial energy
C) Twice the initial energy
D) Thrice the initial energy
Explanation
(d) de-Broglie wavelength
$\lambda = \frac{h}{\sqrt{2mE}}$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{\lambda_{1}}{\lambda_{2}} = \sqrt{\frac{E_{2}}{E_{1}}}\ \ \ \Rightarrow \frac{1 \times 10^{- 9}}{0.5 \times 10^{- 9}} = \sqrt{\frac{E_{2}}{E_{1}}}$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2 = \ \sqrt{\frac{E_{2}}{E_{1}}}\ \ \Rightarrow \frac{E_{2}}{E_{1}} = 4$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ {\ E}_{2}$= 4$E_{1}$
$\therefore$ Energy to be added =$E_{2} - E_{1}$
=$4E_{1} - E_{1} = 3E_{1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.