Question

The energy that should be added to an electron, to reduce its de-broglie wavelength from $2\times {10}^{-9}m$ to $0.5\times {10}^{-9}m$ will be:

• A. $1.1MeV$
• B. $0.56MeV$
• C. $0.56KeV$
• D. $5.67eV$

Answer 1

The correct option is D $5.67eV$

Given,

$m=9.1\times {10}^{-31}kg$

$h=6.6\times {10}^{-34}$

${\lambda }_1=0.5\times {10}^{-9}m$

${\lambda }_2=2\times {10}^{-9}m$

Wavelength, $\lambda =\frac{h}{p}=\frac{h}{\sqrt{2mE}}$

$E=\frac{h^2}{2m{\lambda }^2}$

$\Delta E=\frac{h^2}{2m}(\frac{1}{{\lambda }_1^2}-\frac{1}{{\lambda }_2^2})$

$\Delta E=\frac{(6.6\times {10}^{-34}{)}^2}{2\times 9.1\times {10}^{-31}}(\frac{1}{(0.5\times {10}^{-9}{)}^2}-\frac{1}{(2\times {10}^{-9}{)}^2})\frac{1}{1.6\times {10}^{-19}}eV$

$\Delta E=5.67eV$

The correct option is D.