Question

An electron has kinetic energy $2.8\times {10}^{-23}J$ de-Broglie wavelength will be nearly $m_e=9.1\times {10}^{-31}kg$

• A.
• B.
• C.
• D.

Answer 1

The correct option is C

Formula for de-Broglie wavelength is

$\lambda =\frac{h}{p}Or\lambda =\frac{h}{mv}\Rightarrow eV=\frac{1}{2}m{v}^{2}orv=\sqrt{\frac{2eV}{m}}$

$\lambda =\frac{h}{\sqrt{2meV}}=$ $\frac{6.62\times {10}^{-34}}{\sqrt{2\times 9.1\times {10}^{-31}\times 2.8\times {10}^{-23}}}$

$\lambda =9.28\times {10}^{-8}meter.$