The transition from the state n = 4 to n =3 in a hydrogen like atom results in ultraviolet radiation. Infrared radiation will be obtained in the transition from

$\begin{array}{l}\frac{1}{\mathrm{\lambda }}\mathrm{\alpha }\left(\frac{1}{{\mathrm{n}}_1^2}-\frac{1}{{\mathrm{n}}_2^2}\right);\therefore \frac{1}{{\mathrm{\lambda }}_{\mathrm{uv}}}\mathrm{\alpha }\left(\frac{1}{3^2}-\frac{1}{4^2}\right)\\ \therefore {\mathrm{\lambda }}_{\mathrm{uv}}\mathrm{\alpha }20.5\end{array}$
Accroding to options
1) $\frac{1}{{\mathrm{\lambda }}_1}\mathrm{\alpha }\left(\frac{1}{2^2}-\frac{1}{3^2}\right)\therefore {\mathrm{\lambda }}_1\mathrm{\alpha }7.2$
2) $\frac{1}{{\mathrm{\lambda }}_2}\mathrm{\alpha }\left(\frac{1}{2^2}-\frac{1}{4^2}\right)\therefore {\mathrm{\lambda }}_2\mathrm{\alpha }5.3$
3) $\frac{1}{{\mathrm{\lambda }}_3}\mathrm{\alpha }\left(\frac{1}{4^2}-\frac{1}{5^2}\right)\therefore {\mathrm{\lambda }}_3\mathrm{\alpha }44.4$
4) $\frac{1}{{\mathrm{\lambda }}_4}\mathrm{\alpha }\left(\frac{1}{1^2}-\frac{1}{2^2}\right)\therefore {\mathrm{\lambda }}_4\mathrm{\alpha }1.3$
For IR radiation :
${\mathrm{\lambda }}_{\mathrm{IR}}>{\mathrm{\lambda }}_{\mathrm{uv}};\mathrm{i}.\mathrm{e}., {\mathrm{\lambda }}_3>{\mathrm{\lambda }}_{\mathrm{uv}}$
So, transition should take place from 5→4