Explanation:
B Given that,
$\mathrm{R}_{\mathrm{H}}=1 \times 10^{7} \mathrm{~m}^{-1}$
Then minimum wavelength of Balmer series,
$\mathrm{n}_{1}=2 \text { and } \mathrm{n}_{2}=\infty$
$\frac{1}{\lambda_{\text {min }}}=\mathrm{R}_{\mathrm{H}}\left(\frac{1}{2^{2}}-\frac{1}{\infty^{2}}\right)=\frac{\mathrm{R}_{\mathrm{H}}}{4}$
$\lambda_{\text {min }}=\frac{4}{\mathrm{R}_{\mathrm{H}}}$
For maximum wavelength of Balmer,
$\mathrm{n}_{1}=2 \text { and } \mathrm{n}_{2}=3$
Then, $\frac{1}{\lambda_{\max }}=\mathrm{R}_{\mathrm{H}}\left(\frac{1}{2^{2}}-\frac{1}{3^{3}}\right)=\mathrm{R}_{\mathrm{H}}\left(\frac{1}{4}-\frac{1}{9}\right)=\frac{5 \mathrm{R}_{\mathrm{H}}}{36}$
$\lambda_{\text {max }}=\frac{36}{5 R_{H}}$
Then, difference $\lambda_{\text {max }}-\lambda_{\text {min }}=\frac{36}{5 R_{H}}-\frac{4}{R_{H}}$
$\Delta \lambda=\frac{4}{\mathrm{R}_{\mathrm{H}}}\left(\frac{9}{5}-1\right)=\frac{4}{\mathrm{R}_{\mathrm{H}}} \times \frac{4}{5}$
$\Delta \lambda=\frac{16}{1 \times 10^{7} \times 5}=3.2 \times 10^{-7} \mathrm{~m}^{-2}$
$\Delta \lambda=3200 \AA$