The wave number of the last line of the Balmer series in hydrogen spectrum will be (Rydberg's constant…
The wave number of the last line of the Balmer series in hydrogen spectrum will be (Rydberg's constant $=10^7 \mathrm{~m}^{-1}$ )
$2.5 \times 10^6 \mathrm{~m}^{-1}$
$0.255 \times 10^9 \mathrm{~m}^{-1}$
$250 m^{-1}$
$2.5 \times 10^5 \mathrm{~m}^{-1}$
Solution
We know that: $\frac{1}{\lambda}=R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$
For last line Balmer's series, $n_1=2, n_2=\propto$
So, $\frac{1}{\lambda}=10^7\left(\frac{1}{2^2}-\frac{1}{\infty^2}\right)=0.25 \times 10^7 \mathrm{~m}^{-1}$