Question

The wavelength of first line of Balmer spectrum of hydrogen will be :

• A. $4340\mathring{A}$
• B. $4101\mathring{A}$
• C. $6569\mathring{A}$
• D. $4861\mathring{A}$

Answer 1

Answer: (C)

$\frac{1}{\lambda }=R_H\left[\frac{1}{2^2}-\frac{1}{n^2}\right]$

for first line $n=3$,

$\therefore \frac{1}{\lambda }=109678\left[\frac{1}{2^2}-\frac{1}{3^2}\right]$

$\lambda =6569\mathring{A}$