Question

In hydrogen atom, the de-Broglie wavelength of an electron in the second Bohr orbit is
[Given that, Bohr radius, $a_{0}=52.9\, pm$]

• A. $211.6\, pm$
• B. $211.6\, \pi\, pm$
• C. $52.9\, \pi\, pm$
• D. $105.8\, pm$

Answer 1

Answer: (B)

According to Bohr,
$m v r=\frac{n h}{2 \pi}$
$2 \pi r=\frac{n h}{m v}=n \lambda$ ...(i)
$\left[\because \lambda=\frac{h}{m v}\right]$
where, $r=$ radius, $\lambda=$ wavelength
$n=$ number of orbit
Also, $r=\frac{a_{0} n^{2}}{Z}$ ...(ii)
where, $a_{0}=$ Bohr radius $=52.9\, pm$
$Z=$ atomic number
On substituting the value of ' $r$ ' from Eq. (ii) to Eq. (i), we get
$n \lambda =\frac{2 \pi n^{2} a_{0}}{Z}$
$\Rightarrow \lambda=\frac{2 \pi n a_{0}}{Z}$
$\lambda =2 \pi \times 2 \times 52.9$
$[\because n=2, Z=1]$
$=211.6\,\pi\, pm$