Question

If the wavelength of first member of Balmer series of hydrogen spectrum is 6564 A$^\circ$, the wavelength of second member of Balmer series will be:

• A. 1215 $A^\circ$
• B. 4848 $A^\circ$
• C. 6050 $A^\circ$
• D. data given is insufficient to calculate the value

Answer 1

Answer: (B)

4848 $A^\circ$

Answer 2

hint: $\frac{1}{\lambda} = RZ^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$ for balmer series $n_1 =2$ and $n_2 = 3,4,5.....$