Question

The $\lambda$ for $H_{\alpha }$ line of Balmer series is 6500 $\mathring{A}$. Thus, $\lambda$ for $H_{\beta }$ line of Balmer series is:
(a) $4814\mathring{A}$ (b) $4914\mathring{A}$
(c) $5014\mathring{A}$ (d) $4714\mathring{A}$

Answer 1

The A option is correct

for Ha line $\to n_1=2$ and $n_2$=3

for Hb line$\to n_1=2$ and $n_2=4$

$\frac{1}{{\alpha }_a}=R_h[\frac{1}{2^2}-\frac{1}{3^2}]$...........1

$\frac{1}{{\alpha }_b}=R_h[\frac{1}{2^2}-\frac{1}{4^2}]$...........2

from 1 and 2

$\frac{{\alpha }_b}{{\alpha }_a}=\frac{16\times 5}{36\times 3}$

${\alpha }_b={\alpha }_a\frac{80}{108}$=6500\times \frac{80}{108}$

${\alpha }_b=4814.8A^0$