Q175 / 180 · KCET 2010 · Physics
The de-Broglie wavelength of the electron in the ground state of the hydrogen atom is ..... (radius of the first orbit of hydrogen atom $=0.53 Å$ ).
- A$1.67 \mathrm{~A}$
- Correct answerB$3.33 Å$
- C$1.06 Å$
- D$0.53 Å$
Solution
According to Bohr's quantisation of angular momentum
$$
m v r=\frac{n h}{2 \pi}
$$
or
$\frac{\mathrm{h}}{\mathrm{mv}}=\frac{2 \pi \mathrm{r}}{\mathrm{n}} \quad \text{...(i)}$
de- Broglie wavelength
$$
\lambda=\frac{\mathrm{h}}{\mathrm{mv}} \quad \text{...(ii)}
$$
From Eqs. (i) and (ii), we get
Wavelength $\lambda=\frac{2 \pi \mathrm{r}}{\mathrm{n}}$
$$
\begin{aligned}
&=\frac{2 \times \pi \times 0.53 Å}{1} \\
&=3.33 Å
\end{aligned}
$$
Hence, the correct answer is option: 'B'.