Question

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

Answer 1

The correct option is D Thrice the initial energy

$\lambda =\frac{h}{\sqrt{2mE}}$

so $\frac{{\lambda }^{\prime }}{\lambda }=\frac{\sqrt{E}}{E^{\prime }}$

$\therefore \frac{E}{E^{\prime }}=(\frac{{\lambda }^{\prime }}{\lambda }{)}^2$

$=(\frac{0.5}{1}{)}^2=\frac{1}{4}$

so E' = 4E

so difference is 3E