Question

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

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

Answer 1

The correct option is B $211.6\pi \text{pm}$

Bohr radius, $a_0=52.9\text{pm}$
$n=2,r_n=n^2{a}_0=(2{)}^2{a}_0=4\times 52.9\text{pm}=211.6\text{pm}$. The angular momentum of an electron in a given stationary state can be expressed as in equation,
$mvr=n\frac{h}{2\pi }\Rightarrow mvr=2\frac{h}{2\pi }=\frac{h}{\pi }\Rightarrow mvr\pi =h...\left(i\right)$
We know that,
De broglie equation is given as,
$\lambda =\frac{h}{mv}$
$\Rightarrow \lambda mv=h...\left(ii\right)$
From equation $\left(i\right)$ and $\left(ii\right)$.
We get,
$\lambda =\pi r$
Putting the value of $r$,
$\lambda =211.6\pi \text{pm}$