Question

The transition from the state $n=3$ to $n=1$in a hydrogen like atom results in ultraviolet radiation. Infrared radiation will be obtained in the transition from:
$(a)3\to 2$
$(b)4\to 2$
$(c)4\to 3$
$(d)2\to 1$

Answer 1

We will make use of Spectral Series of Hydrogen or Hydrogen-like atom in the above question as under proper radiation, these atoms undergo excitation followed by de-excitation which results in the release of photons, thus light waves of different wavelength being emitted from the atom.

Complete answer:
We know that, in the Line Spectrum Series of Hydrogen or hydrogen-like atoms, the electrons in the atoms when kept under radiation excite to higher orbital levels. They then de-excite to come back to a more stable inner orbital. On de-excitation these electrons release photons of different wavelengths that further give different spectrum series. These are:
Lyman series $(n\to 1)$ is the ultraviolet region.
Balmer series $(n\to 2)$is the visible spectrum.
Paschen series $(n\to 3)$is the infrared region.
Thus, in a hydrogen like atom, a transition from $(4\to 3)$, i.e., option $(c)$ will result in Infrared radiation as it lies in the Paschen series.
In options $(a)$and $(b)$, the transition is from $(n\to 2)$, i.e., the Balmer series of light, thus they lie in the visible spectrum.
And in option $(d)$, the transition is from $(n\to 1)$, thus it lies in the ultraviolet region of the Lyman series.

Hence, option $(c)$ is the only correct option.

Note:
The Bracket $(n\to 4)$ , Pfund $(n\to 5)$ and Humphreys $(n\to 6)$ series also lie in the Infrared region. So, we should always check if more than one option is correct. Also, we shouldn’t misinterpret one series with another, so it is always important to remember these concepts properly, so that we don’t get confused while solving a problem.