Question

If the wavelength of light that is emitted from hydrogen atom when an electron falls from orbit $n=2$ to orbit $n=1$ is $122nm$, then minimum wavelength of the series is:

• A. $9150\mathring{A}$
• B. $812\mathring{A}$
• C. $915\mathring{A}$
• D. $405\mathring{A}$

Answer 1

The correct option is C $915\mathring{A}$

Given : $\lambda =122nm$ for $n_1=1$ and $n_2=2$

$\therefore$ $\frac{1}{\lambda }=R(\frac{1}{n_1^2}-\frac{1}{n_2^2})$

$⟹$ $\frac{1}{122}=R(\frac{1}{1^2}-\frac{1}{2^2})=\frac{3R}{4}$ ...............(1)

Wavelength is minimum when $n_1=1$ and $n_2=\infty$

$⟹$ $\frac{1}{{\lambda }_{min}}=R(\frac{1}{1^2}-\frac{1}{\infty })=R$ ................(2)

Dividing (1) and (2) we get: $\frac{{\lambda }_{min}}{122}=\frac{3}{4}$

$⟹$ ${\lambda }_{min}=0.75\times 122=91.5nm$ $=915$ $A^o$