Question

An electron with an initial kinetic energy of $100eV$ is accelerated through a potential difference of $50V$. 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 $1\stackrel{o}{A}$

The initial kinetic energy of the electron is $100eV$

Now, it is accelerated through a potential is $50V$

Therefore, the additional kinetic energy it gains is

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

Hence, the total kinetic energy possessed by the electron is $150eV$ which is equivalent to $150\times 1.6\times {10}^{-19}J$

$\lambda =\frac{h}{\sqrt{2mK}}=\frac{6.63\times {10}^{-34}}{\sqrt{2\times 9.1\times {10}^{-31}\times 150\times 1.6\times {10}^{-19}}}=1\times {10}^{-10}m=1\stackrel{o}{A}$