Question

What is the De-Broglie wavelength associated with the hydrogen electron in its third orbit :

• A. $9.96\times {10}^{-10}cm$
• B. $9.96\times {10}^{-8}cm$
• C. $9.96\times {10}^{4}cm$
• D. $9.96\times {10}^{8}cm$

Answer 1

The correct option is B $9.96\times {10}^{-8}cm$

Radius of 3rd orbit $r=R\times 3^2=9R$

Since,

$\frac{nh}{2\pi }=mvr$

$\frac{h}{mv}=\frac{2\pi r}{n}$

$\lambda =\frac{h}{mv}=\frac{2\pi \times 9R}{3}$

$\lambda =6\pi R$

We know that

$R=0.529A^0$

Therefore,

$\lambda =6\pi \times 0.529$

$\lambda =9.96A^0$

$\lambda =9.96\times {10}^{-8}cm$

Hence, this is the answer.