NEET UG 2019 — Chemistry Physical Chemistry

2019mcqhard

In hydrogen atom, the de-Broglie wavelength of an electron in the second Bohr orbit is [Given that, Bohr radius, $\mathrm{a}_0=52.9 \mathrm{pm}$ ]

Official previous-year question

Held on 30 Apr 2019 · Verified 9 Jul 2026.

Options

A
$211.6 \mathrm{pm}$
B
$211.6 \pi \mathrm{pm}$
C
$52.9 \pi \mathrm{pm}$
D
$105.8 \mathrm{pm}$

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Solution

According to Bohr, $\begin{aligned} & \mathrm{mvr}=\frac{\mathrm{nh}}{2 \pi} \\ & 2 \pi \mathrm{r}=\frac{\mathrm{nh}}{\mathrm{mv}}=\mathrm{n} \lambda \end{aligned}$ ..(i) $\left[\because \lambda=\frac{\mathrm{h}}{\mathrm{mv}}\right]$ where, $\mathrm{r}=$ radius, $\lambda=$ wavelength $\mathrm{n}=$ number of orbit Also, $r=\frac{a_0 n^2}{\mathrm{Z}}$ where, $\mathrm{a}_0=$ Bohr radius $=52.9 \mathrm{pm}$ $\mathrm{Z}=$ atomic number On substituting the value of ' $r$ ' from Eq. (ii) to Eq. (i), we get $\begin{aligned} \mathrm{n} \lambda & =\frac{2 \pi \mathrm{n}^2 \mathrm{a}_0}{\mathrm{Z}} \Rightarrow \lambda=\frac{2 \pi \mathrm{na}_0}{\mathrm{Z}} \\ \lambda & =2 \pi \times 2 \times 52.9 \quad[\because \mathrm{n}=2, \mathrm{Z}=1] \\ & =211.6 \pi \mathrm{pm} \end{aligned}$

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NEET UG 2019 — Chemistry Physical Chemistry

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NEET UG 2019 — Chemistry Physical Chemistry

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