Question

An electron with initial kinetic energy of $100$eV is acceleration through a potential difference of $50$V. Now the de-Broglie wavelength of electron becomes.

• A. $1\stackrel{o}{A}$
• B. $\sqrt{1.5}\stackrel{o}{A}$
• C. $\sqrt{3}\stackrel{o}{A}$
• D. $12.27\stackrel{o}{A}$

Answer 1

The correct option is A $1\stackrel{o}{A}$

Additional kinetic energy it gains is

$K^{\prime }=eV=50eV$

Total kinetic energy is equal to $150\times 1.6\times {10}^{-19}J$

The de broglie wavelength is

$\lambda =\frac{h}{\sqrt{2mK}}=\frac{6.63\times {10}^{-34}}{\sqrt{2\times 9.1\times {10}^{-34}\times 150\times 1.6\times {10}^{-19}}}=1\times {10}^{-10}m$
$\lambda =1A^o$