Question

The frequency of light emitted for the transition n = 4 to n = 2 of $H{e^ + }$ is equal to the transition in H atom corresponding to which of the following?
(A) n = 3 to n = 1
(B) n = 2 to n = 1
(C) n = 3 to n = 2
(D) n = 4 to n = 3

Answer 1

Hint: For any atom, the energy change associated with the transition of electrons is given by
\[\Delta E = hv = \dfrac{{2{\pi ^2}m{Z^2}{e^4}{k^2}}}{{{h^2}}}\left[ {\dfrac{1}{{{n_1}^2}} - \dfrac{1}{{{n_2}^2}}} \right]\]
Where Z=Atomic number of the atom

Complete step by step solution:
We will first find the frequency related to the given transition in $H{e^ + }$ atom and then compare it with the given transitions for the H atom.
We know that $\Delta E = hv$
Where $v$=frequency and h= Planck’s constant
Now, in electronic transitions, we can give the energy change by the following equation.
\[\Delta E = hv = \dfrac{{2{\pi ^2}m{Z^2}{e^4}{k^2}}}{{{h^2}}}\left[ {\dfrac{1}{{{n_1}^2}} - \dfrac{1}{{{n_2}^2}}} \right]\] ………………..(1)
Here, electrons fall from ${n_2}$ level to ${n_1}$ level.
Now, in the case of $H{e^ + }$, we are given that ${n_2} = 4$ and ${n_1} = 2$. If we take the value of $\dfrac{{2{\pi ^2}m{e^4}{k^2}}}{{{h^3}}}$ as a constant, then we can write equation (1) as
\[{v_{H{e^ + }}} = {\text{constant}} \times {{\text{Z}}^2}\left[ {\dfrac{1}{{{n_1}^2}} - \dfrac{1}{{{n_2}^2}}} \right]\]
Here, Z= atomic number of the atom and for $H{e^ + }$, it is 2.
So, we can write that
\[{v_{H{e^ + }}} = {\text{constant}} \times {2^2}\left[ {\dfrac{1}{{{2^2}}} - \dfrac{1}{{{4^2}}}} \right]\]
\[{v_{H{e^ + }}} = {\text{constant}} \times 4\left[ {\dfrac{1}{4} - \dfrac{1}{{16}}} \right]\]
\[{v_{H{e^ + }}} = {\text{constant}} \times \dfrac{3}{4}\]
Now, we will see the same for H-atom. We can write the frequency of H-atom will be
\[{v_H} = {\text{constant}} \times {{\text{Z}}^2}\left[ {\dfrac{1}{{{n_1}^2}} - \dfrac{1}{{{n_2}^2}}} \right]\]
For n=3 to n=1,
\[{v_H} = {\text{constant}} \times {1^2}\left[ {\dfrac{1}{{{1^2}}} - \dfrac{1}{{{3^2}}}} \right]{\text{ }}\because {\text{Z = 1}}\]
\[{v_H} = {\text{constant}} \times \left[ {1 - \dfrac{1}{9}} \right] = {\text{constant}} \times \dfrac{8}{9}\]
For n=2 to n=1
\[{v_H} = {\text{constant}} \times {1^2}\left[ {\dfrac{1}{{{1^2}}} - \dfrac{1}{{{2^2}}}} \right]\]
\[{v_H} = {\text{constant}} \times \left[ {1 - \dfrac{1}{4}} \right] = {\text{constant}} \times \dfrac{3}{4}\]
For n=3 to n=2
\[{v_H} = {\text{constant}} \times {1^2}\left[ {\dfrac{1}{{{2^2}}} - \dfrac{1}{{{3^2}}}} \right]\]
\[{v_H} = {\text{constant}} \times \left[ {\dfrac{1}{4} - \dfrac{1}{9}} \right] = {\text{constant}} \times \dfrac{5}{{36}}\]
For n=4 to n=3,
\[{v_H} = {\text{constant}} \times {1^2}\left[ {\dfrac{1}{{{3^2}}} - \dfrac{1}{{{4^2}}}} \right]\]
\[{v_H} = {\text{constant}} \times \left[ {\dfrac{1}{9} - \dfrac{1}{{16}}} \right] = {\text{constant}} \times \dfrac{7}{{144}}\]
So, if we compare the frequency of $H{e^ + }$ for the given transition and H-atom, they are the same if the transition is from n=2 to n=1 in H-atom.

Therefore correct answer is (B).

Note: Note that here charge on the atom will not make any difference to the energy or frequency related to the transition of the electron. So, do not assume the Z is the charge on the atom. Actually, it is the atomic number of the atom in the equation.