Question

The wavelength of the first number of Balmer series of hydrogen spectrum is:

[R for hydrogen = 1.09678 $\times$ 10${}^7$ m${}^{-1}$]

• A. $6564\mathring{A}$
• B. $5743\mathring{A}$
• C. $3782\mathring{A}$
• D. $7532\mathring{A}$

Answer 1

The correct option is A $6564\mathring{A}$

Balmer series:

$\frac{1}{\lambda }=R\left[\frac{1}{2^2}-\frac{1}{n^2}\right]$, n = 3, 4, 5,...

When n = 3 (first member of Balmer series),

$\frac{1}{\lambda }=1.09678\times {10}^7\left(\frac{1}{2^2}-\frac{1}{3^2}\right)$

$=1.09678\times {10}^7\times \frac{5}{36}$

$\therefore \lambda =6564\times {10}^{-10}m=6564\mathring{A}$

Hence, option A is correct.