Question

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
(A) $2 \to 1$
(B) $3 \to 2$
(C) $4 \to 2$
(D) $5 \to 4$

Answer 1

To solve this question, we need to use Rydberg's formula, with the help of which we can make the comparison between different transitions, and relate it with the energies of the emitted radiations. We can take the help of the electromagnetic spectrum to compare the energies of different radiations.

Formula used:
$\Rightarrow \dfrac{1}{\lambda } = R\left( {\dfrac{1}{{{n_i}^2}} - \dfrac{1}{{{n_f}^2}}} \right)$, where $\lambda$ is the wavelength of the radiation emitted when an electron returns from the state ${n_f}$ to the state ${n_i}$.

Complete step by step solution:
The energy of an electromagnetic wave, as we know is given by
$\Rightarrow E = h\nu$
We know that $\nu = \dfrac{c}{\lambda }$. So we have
$\Rightarrow E = \dfrac{{hc}}{\lambda }$
Since $h$ and $c$are constants, therefore
$\Rightarrow E \propto \dfrac{1}{\lambda }$ …………….(i)
Now, for hydrogen like atom we have the inverse of the wavelength of the emitted radiation given by
$\Rightarrow \dfrac{1}{\lambda } = R\left( {\dfrac{1}{{{n_i}^2}} - \dfrac{1}{{{n_f}^2}}} \right)$ ……...(ii)
As the wavelength is always positive, so the RHS of the above expression is positive, that is,
$\Rightarrow \left( {\dfrac{1}{{{n_i}^2}} - \dfrac{1}{{{n_f}^2}}} \right) > 0$
From (i) and (ii) we have
$\Rightarrow E \propto R\left( {\dfrac{1}{{{n_i}^2}} - \dfrac{1}{{{n_f}^2}}} \right)$
Since$R$is a constant, so we have
$\Rightarrow E \propto \left( {\dfrac{1}{{{n_i}^2}} - \dfrac{1}{{{n_f}^2}}} \right)$ ……………..(iii)
Now, from the electromagnetic spectrum we have the following order of the energies
$\Rightarrow {\text{Infrared}} < {\text{Ultraviolet}}$
So the energy of the infrared radiation is less than that of the ultraviolet radiation.
As we can see from the above expression the energy of the emitted radiation will decrease with the increase in the value of ${n_i}$. So, for the infrared radiation, which has less energy than the energy of the ultraviolet radiation, it should have the ${n_i}$ value greater than that of the ultraviolet radiation. According to the question, the ultraviolet radiation result from the transition $n = 4$ to $n = 3$
so the value of ${n_i}$ for the ultraviolet radiation is equal to $4$. Looking in the options, we find the only transition $5 \to 4$ which has ${n_i}$ value equal to $5$, which is greater than $4$.
Hence, the correct answer is option (D).

Note:
Do not try to obtain the exact value of the wavelength of the infrared radiation from the formula given in the solution. The comparison method always works for these types of questions, and it is the most efficient method.