Question

The de-Broglie wavelength of the electron in the ground state of the hydrogen atom is,
(radius of the first orbit of hydrogen atom $=0.53\mathring{\text{A}}$)

• A. $1.67\mathring{\text{A}}$
• B. $3.33\mathring{\text{A}}$
• C. $1.06\mathring{\text{A}}$
• D. $0.53\mathring{\text{A}}$

Answer 1

The correct option is B $3.33\mathring{\text{A}}$

According to Bohr's quantization condition,

$mvr=\frac{nh}{2\pi }\Rightarrow \frac{h}{mv}=\frac{2\pi r}{n}......\left(i\right)$

de-Broglie wavelength of an electron is,
$\lambda =\frac{h}{mv}......\left(ii\right)$

Equating $\left(i\right)$ and $\left(ii\right)$, we get,

$\lambda =\frac{2\pi r}{n}$

For the first orbit, $r=0.53\mathring{\text{A}},n=1$

$\therefore \lambda =\frac{2\times 3.14\times 0.53}{1}=3.33\mathring{\text{A}}$

Hence, $\left(B\right)$ is the correct answer.