Q42 / 180 · NEET 2019 (Odisha) · Chemistry

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

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

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 $$ \begin{aligned} \mathrm{n} \lambda & =\frac{2 \pi \mathrm{n}^2 \mathrm{a}_0}{\mathrm{Z}} \Rightarrow \lambda=\frac{2 \pi \mathrm{na}_0}{\mathrm{Z}} \\ \lambda & =2 \pi \times 2 \times 52.9 \quad[\because \mathrm{n}=2, \mathrm{Z}=1] \\ & =211.6 \pi \mathrm{pm} \end{aligned} $$

Hence, the correct answer is option: 'B'.

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