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

The energy of a photon emitted when an electron in a hydrogen-like atom jumps from level $$n_i$$ to $$n_f$$ is

$$E = 13.6\,Z^2\Bigl(\frac{1}{n_f^{\,2}}-\frac{1}{n_i^{\,2}}\Bigr)\,\text{eV}$$

where $$Z$$ is the atomic number (for hydrogen, $$Z = 1$$).
The wavelength is related to energy by $$E = \frac{hc}{\lambda}$$, so

• larger $$\Bigl(\frac{1}{n_f^{\,2}}-\frac{1}{n_i^{\,2}}\Bigr) \;\Rightarrow\;$$ larger energy $$E$$ and hence smaller wavelength (towards ultraviolet),
• smaller $$\Bigl(\frac{1}{n_f^{\,2}}-\frac{1}{n_i^{\,2}}\Bigr) \;\Rightarrow\;$$ smaller energy $$E$$ and hence larger wavelength (towards infrared).

The question says that the jump $$4 \to 3$$ produces ultraviolet (UV) radiation. For each offered transition, compute the factor $$\Delta=\frac{1}{n_f^{\,2}}-\frac{1}{n_i^{\,2}}$$ and compare it with the UV-producing value.

For the given UV transition: $$4 \to 3$$

$$\Delta_{UV} = \frac{1}{3^{2}} - \frac{1}{4^{2}} = \frac{1}{9} - \frac{1}{16} = \frac{7}{144}\; (\approx 0.0486)$$

Now evaluate each option.

$$\Delta = 1 - \frac{1}{4} = \frac{3}{4} \approx 0.75$$ (much larger than 0.0486 ⇒ far-UV)

$$\Delta = \frac{1}{2^{2}} - \frac{1}{3^{2}} = \frac{1}{4} - \frac{1}{9} = \frac{5}{36} \approx 0.139$$ (larger than 0.0486 ⇒ still higher-energy, around visible/near-UV)

$$\Delta = \frac{1}{2^{2}} - \frac{1}{4^{2}} = \frac{1}{4} - \frac{1}{16} = \frac{3}{16} = 0.1875$$ (even larger ⇒ UV)

$$\Delta = \frac{1}{4^{2}} - \frac{1}{5^{2}} = \frac{1}{16} - \frac{1}{25} = \frac{9}{400} = 0.0225$$

This $$\Delta$$ is about half that of the $$4 \to 3$$ jump, giving roughly half the energy and therefore roughly twice the wavelength, moving the radiation past the visible range into the infrared.

Hence the transition that yields infrared radiation is:

Option D which is: $$5 \to 4$$