Question

In the spectrum of singly ionised helium, the wavelength of a line observed is almost the same as the first line of the Balmer series of hydrogen. It is due to transition of electron
A. from ${n_1} = 6$ to ${n_2} = 4$
B. from ${n_1} = 5$ to ${n_2} = 3$
C. from ${n_1} = 4$ to ${n_2} = 2$
D. from ${n_1} = 3$ to ${n_2} = 2$

Answer 1

Hint:the wavelength of the emitted spectrum of hydrogen can be calculated using the formulae $\dfrac{1}{\lambda } = {R_H}{Z^2}(\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_2^2}})$. The same formulae can be written for the emitted spectrum of helium ions. Since both wavelengths are said to be equal the two values can be equated to obtain the relation between principal quantum numbers of helium.

Complete step by step answer:
The emission spectrum of hydrogen has been divided into different series based on the lower energy level to which the electron transitions are taking place. Balmer series in the name given for the spectral emissions of hydrogen which arises due to the transition from higher energy levels to the energy level with principal quantum number 2. Balmer emission spectrum lies in the range of $383.5384\,nm$ to $656.2852\,nm$. As we know this spectrum lies in the region of visible spectrum, and thus Balmer emission is visible to naked eyes.

We can predict the wavelength of the emission spectrum if we know the initial and final principal quantum number of the energy levels in which the transmission took place.
The relation can be mathematically plotted as
$\dfrac{1}{\lambda } = {R_H}{Z^2}(\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_2^2}})$
${R_H}$ represents the Rydberg constant, and the value is
${R_H} = 1.09677576 \times {10^7}{m^{ - 1}}$for hydrogen.
The value of Rydberg constant varies from element to element, Rydberg constant for an element can be expressed as
${R_H} = \dfrac{{m{e^4}}}{{8{h^3}c{\varepsilon _0}}}$

Now we can move back to our question,
It was said the wavelength of helium (atomic number 2) and the wavelength of the first line of Balmer 5series hydrogen (atomic number 1) was the same.
Mathematically,
${\lambda _H} = {\lambda _{He}}$
$ \Rightarrow \dfrac{1}{{{\lambda _H}}} = \dfrac{1}{{{\lambda _{He}}}}$
$ \Rightarrow {1^2}(\dfrac{1}{{{2^2}}} - \dfrac{1}{{{3^2}}}) = {2^2}(\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_{2}^2}})$
$(\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_{2}^2}}) = \dfrac{5}{{36}}$
Thus, we have successfully found the value of $(\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_{2}^2}})$. now we have to choose the value of principal quantum number in a manner such that it satisfies the above equation.If we consider the first option ${n_1} = 6$and ${n_2} = 4$.
We get $(\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_{2}^2}}) = (\dfrac{1}{{{6^2}}} - \dfrac{1}{{{4^2}}}) = - \dfrac{5}{{36}}$
Remember that the negative sign indicates energy released. Therefore, considering the positive integral solutions ${n_1} = 6$and ${n_2} = 4$

Thus, option A is the correct answer.

Note:The value of $(\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_{2}^2}})$ will be negative as it denotes the energy released. Getting a negative sign doesn’t mean here that you got the wrong answer, but the nature of energy released. Multiplying with the value of Rydberg constant wastes the time, as it gets cancelled later.