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)

Bohr radius, $a_0=52.9pm$ $n=2,r_n=n^2{a}_0=(2{)}^2{a}_0=4\times 52.9pm=211.6pm$ The angular momentum of an electron in a given stationary state can be expressed as in equation, $mvr=n\cdot \frac{h}{2\pi }=2\times \frac{h}{2\pi }=\frac{h}{\pi }$
de-Broglie equation mvr $\pi =h$ $\lambda =\frac{h}{mv};\hat{\lambda }mv=h$
From equatisns, (i) and (ii), we get $\lambda =\pi r$
Putting the value of $r,\lambda =211.6\pi pm$