Q152 / 200 · WBJEE 2013 · Physics
The de-Broglie wavelength of an electron (mass $=1 \times 10^{-30} \mathrm{kg},$ charge $\left.=16 \times 10^{-19} \mathrm{C}\right)$ with a kinetic energy of $200 \mathrm{eV}$ is (Planck's constant $=6.6 \times 10^{-34} \mathrm{Js}$
- A$9.60 \times 10^{-11} \mathrm{m}$
- Correct answerB$8.25 \times 10^{-11} \mathrm{m}$
- C$6.25 \times 10^{-11} \mathrm{m}$
- D$5.00 \times 10^{-11} \mathrm{m}$
Solution
The de-Broglie wavelength, $\lambda=\frac{h}{\sqrt{2 m k}}$
Given,
$$
h=6.6 \times 10^{-34} \mathrm{J}-\mathrm{s}
$$
$$
\begin{aligned}
m &=1 \times 10^{-30} \mathrm{kg} \\
K=200 \mathrm{eV} &=200 \times 1.6 \times 10^{-19} \mathrm{J}
\end{aligned}
$$
Substituting all these values
$$
\begin{aligned}
\lambda &=\frac{6.6 \times 10^{-34}}{\sqrt{2 \times 1 \times 10^{-30} \times 200 \times 1.6 \times 10^{-19}}} \\
&=0.825 \times 10^{-10}=8.25 \times 10^{-11} \mathrm{m}
\end{aligned}
$$
Hence, the correct answer is option: 'B'.