Question

In hydrogen atom, the de-Broglie wavelength of an electron in the second Bohr orbit is
[Given that, Bohr radius, $a_0=52.9pm$]

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

Answer 1

Answer: (B)

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