Q119 / 155 · WBJEE 2016 · Physics
The potential difference $V$ required for accelerating an electron to have the de-Broglie wavelength of $1 Å$ is
- A$100 \mathrm{V}$
- B$125 \mathrm{V}$
- Correct answerC$150 \mathrm{V}$
- D$200 \mathrm{V}$
Solution
The de-Broglie wavelength associated with electron is given by.
$$
\lambda=\frac{h}{\sqrt{2 m e V}}
$$
where, $h=$ Planck's constant
$$
\begin{aligned}
m &=\text { mass of electron } \\
\mathrm{e} &=\text { charge on electron } \\
\mathrm{V} &=\text { potential difference } \\
\lambda &=1 \dot{\mathrm{A}}=10^{-10} \mathrm{m} \\
\lambda^{2} &=\frac{h^{2}}{2 \mathrm{meV}} \Rightarrow \mathrm{V}=\frac{h^{2}}{2 \mathrm{me} \lambda^{2}} \\
&=\frac{\left(6.6 \times 10^{-34}\right)^{2}}{2 \times 9.1 \times 10^{-31} \times 1.6 \times 10^{-19} \times\left(1 \times 10^{-10}\right)^{2}} \\
&=\frac{6.6 \times 10^{-34} \times 6.6 \times 10^{-34}}{2 \times 9.1 \times 10^{-31} \times 1.6 \times 10^{-19} \times 10^{-20}} \\
&=\frac{6.6 \times 6.6 \times 10^{2}}{2 \times 9.1 \times 1.6}=\frac{4356 \times 10^{2}}{2912}=150 \mathrm{V}
\end{aligned}
$$
Hence, the correct answer is option: 'C'.