Question

Hydrogen $\left( {{}_1{{\text{H}}^1}} \right)$,Deuterium $\left( {{}_1{{\text{H}}^2}} \right)$, singly ionized Helium ${\left( {{}_2{\text{H}}{{\text{e}}^4}} \right)^ + }$ and doubly ionized lithium ${\left( {{}_3L{i^6}} \right)^{ + + }}$ all have one electron around the nucleus. Consider an electron transition from $n = 2$ to $n = 1$. If the wavelengths of emitted radiation are ${\lambda _1}$, ${\lambda _2}$, ${\lambda _3}$, and ${\lambda _4}$ respectively then approximately which one of the following is correct?
(A) \[{\lambda _1} = {\lambda _2} = 4{\lambda _3} = 9{\lambda _4}\,\]
(B) \[{\lambda _1} = 2{\lambda _2} = 3{\lambda _3} = 4{\lambda _4}\,\]
(C) \[4{\lambda _1} = 2{\lambda _2} = 2{\lambda _3} = {\lambda _4}\,\]
(D) \[{\lambda _1} = 2{\lambda _2} = 2{\lambda _3} = {\lambda _4}\,\]

Answer 1

Hint: To solve this question, we need to use the Rydberg’s formula for determining the wavelengths corresponding to the given transition for all of the four elements given. Then we can easily determine the relation between all the four wavelengths.
Formula used: The formula used to solve this question is given by
$\dfrac{1}{\lambda } = R{z^2}\left( {\dfrac{1}{{{n_1}^2}} - \dfrac{1}{{{n_2}^2}}} \right)$, here $\lambda $ is the wavelength of the radiation emitted due to the transition ${n_2} \to {n_1}$ of an electron present in the nucleus of an element having the atomic number of $z$.

Complete step-by-step solution:
We know that the wavelength of the emitted radiation due to the transition of electron is given by the Rydberg’s formula, which is written as
$\dfrac{1}{\lambda } = R{z^2}\left( {\dfrac{1}{{{n_1}^2}} - \dfrac{1}{{{n_2}^2}}} \right)$
According to the question, the given transition of the electron is from $n = 2$ to $n = 1$. So we have ${n_1} = 1$ and ${n_2} = 2$. Substituting these values above, we get
$\dfrac{1}{\lambda } = R{z^2}\left( {\dfrac{1}{{{1^2}}} - \dfrac{1}{{{2^2}}}} \right)$
\[ \Rightarrow \dfrac{1}{\lambda } = \dfrac{3}{4}R{z^2}\]
Taking the reciprocal, we get
$\lambda = \dfrac{4}{{3R{z^2}}}$ (1)
Now, we consider each of the four elements given in the question separately.
For Hydrogen $\left( {{}_1{{\text{H}}^1}} \right)$:
We know that the atomic number of hydrogen is $1$. Therefore substituting $z = 1$ in (1) we get
${\lambda _1} = \dfrac{4}{{3R{{\left( 1 \right)}^2}}}$
$ \Rightarrow {\lambda _1} = \dfrac{4}{{3R}}$ (2)
For Deuterium $\left( {{}_1{{\text{H}}^2}} \right)$:
We know that the atomic number of Deuterium is also $1$. So the wavelength in this case will be equal to that in the previous case, that is,
\[{\lambda _1} = {\lambda _2}\] (3)
For singly ionized Helium ${\left( {{}_2{\text{H}}{{\text{e}}^4}} \right)^ + }$:
Since the atomic number of Helium is equal to $2$. Therefore substituting $z = 2$ in (1) we get
${\lambda _3} = \dfrac{4}{{3R{{\left( 2 \right)}^2}}}$
$ \Rightarrow {\lambda _3} = \dfrac{1}{{3R}}$ (4)
For doubly ionized lithium ${\left( {{}_3L{i^6}} \right)^{ + + }}$:
We know that the atomic number of hydrogen is equal to $3$. Therefore substituting $z = 3$ in (1) we get
${\lambda _4} = \dfrac{4}{{3R{{\left( 3 \right)}^2}}}$
$ \Rightarrow {\lambda _4} = \dfrac{4}{{27R}}$ (5)
Dividing (4) by (2) we get
$\dfrac{{{\lambda _3}}}{{{\lambda _1}}} = \dfrac{1}{{3R}} \times \dfrac{{3R}}{4}$
$ \Rightarrow {\lambda _1} = 4{\lambda _3}$ (6)
Dividing (5) by (2) we get
$\dfrac{{{\lambda _4}}}{{{\lambda _1}}} = \dfrac{4}{{27R}} \times \dfrac{{3R}}{4}$
$ \Rightarrow {\lambda _1} = 9{\lambda _4}$ (7)
From (3), (6) and (7) we can conclude that
\[{\lambda _1} = {\lambda _2} = 4{\lambda _3} = 9{\lambda _4}\,\]

Hence, the correct answer is option A.

Note: We should not forget the factor of the atomic number that appears in the Rydberg’s formula. Rydberg's formula is also famous for the hydrogen atom. But that formula works only for hydrogen and its isotopes. So we should remember the Rydberg’s formula used in this solution, which is the more general formula.