Question

What transition in the hydrogen spectrum would have the same wavelength as the Balmer transition, n=4 to n=2 of \[H{{e}^{+}}\]spectrum?
(a) n = 4 to n = 2
(b) n = 3 to n = 2
(c) n = 2 to n = 1
(d) n = 4 to n = 3

Answer 1

Hint: Electronic transitions occur when an electron in a lower energy level acquires energy and moves to higher energy. When it comes back to lower energy, it emits this energy. In the Balmer series, the transition is from higher energy level to second energy level.

Complete step by step solution:
For solving this question, we need to compare the wavelength of transition in the H-spectrum with the Balmer transition n=4 to n=2 of \[H{{e}^{+}}\]spectrum.
So, we consider \[{{\lambda }_{H}}={{\lambda }_{H{{e}^{+}}}}\],where \[{{\lambda }_{H}}\]wavelength for hydrogen atom and \[{{\lambda }_{H{{e}^{+}}}}\]wavelength for helium ion.
We have to use Rydberg formula for any hydrogen like element, which is
\[\dfrac{1}{\lambda }=R{{Z}^{2}}(\dfrac{1}{n_{1}^{2}}-\dfrac{1}{n_{2}^{2}})\], where \[\lambda \]is the wavelength of the light emitted, R is the Rydberg constant for the element (in our case it is helium atom), Z is the atomic number of helium, \[{{n}_{1}}\] is the principal quantum number of the upper energy level and \[{{n}_{2}}\]is the principal quantum number of the lower energy level in the electronic transition.

So for hydrogen
\[\dfrac{1}{{{\lambda }_{H}}}={{R}_{H}}Z_{H}^{2}(\dfrac{1}{n_{1}^{2}}-\dfrac{1}{n_{2}^{2}})\]
For helium
\[\dfrac{1}{{{\lambda }_{H{{e}^{+}}}}}={{R}_{H{{e}^{+}}}}Z_{H{{e}^{+}}}^{2}(\dfrac{1}{n_{1}^{2}}-\dfrac{1}{n_{2}^{2}})={{R}_{H{{e}^{+}}}}Z_{H{{e}^{+}}}^{2}(\dfrac{1}{2_{{}}^{2}}-\dfrac{1}{4_{{}}^{2}})\],\[{{n}_{1}}\]=2 and \[{{n}_{2}}\]=4

As we discussed above, for hydrogen and helium \[{{\lambda }_{H}}={{\lambda }_{H{{e}^{+}}}}\], we can write
\[{{R}_{H}}Z_{H}^{2}(\dfrac{1}{n_{1}^{2}}-\dfrac{1}{n_{2}^{2}})={{R}_{H}}Z_{H{{e}^{+}}}^{2}(\dfrac{1}{2_{{}}^{2}}-\dfrac{1}{4_{{}}^{2}})\]
Both the Rydberg constant gets cancelled, so we get
\[Z_{H}^{2}(\dfrac{1}{n_{1}^{2}}-\dfrac{1}{n_{2}^{2}})=Z_{H{{e}^{+}}}^{2}(\dfrac{1}{2_{{}}^{2}}-\dfrac{1}{4_{{}}^{2}})\]
For helium, \[Z_{H{{e}^{+}}}^{2}={{2}^{2}}=4\] and for Hydrogen \[Z_{H}^{2}\]=1. Substituting these values, we get

\[1\times (\dfrac{1}{n_{1}^{2}}-\dfrac{1}{n_{2}^{2}})=4\times (\dfrac{1}{2_{{}}^{2}}-\dfrac{1}{4_{{}}^{2}})\]
=\[(\dfrac{1}{n_{1}^{2}}-\dfrac{1}{n_{2}^{2}})=4\times (\dfrac{1}{4}-\dfrac{1}{16})=\dfrac{3}{4}\]--------(i))
Now by trial and error method, if \[{{n}_{1}}\]=1, then \[{{n}_{2}}\] can be 2, 3..
If we consider the first line, \[{{n}_{2}}\]=2 and \[{{n}_{1}}\]=1.
So, \[(\dfrac{1}{n_{1}^{2}}-\dfrac{1}{n_{2}^{2}})=\dfrac{1}{1}-\dfrac{1}{4}=\dfrac{4-1}{4}=\dfrac{3}{4}\]--------(ii)

We can see that equations (i) and (ii) are equal.
Thus, the transition \[{{n}_{2}}\]=2 to \[{{n}_{1}}\]=1 will give the spectrum of the same wavelength as that of Balmer transition n=4 o n=2 of \[H{{e}^{+}}\]spectrum. So the correct option to the question is (c).

Note: We should note that the Rydberg formula for hydrogen and hydrogen like species is different. This difference is in the atomic number. The atomic number of hydrogen is 1, so we don’t include Z in the formula, but in hydrogen like species like helium, we need to include atomic number Z of helium.