Question

What is the wavelength of light emitted when the electron in a hydrogen atom undergoes transition from the energy level with n=4 to energy level n=2? (Given ${R_H} = 109678\,{\rm{c}}{{\rm{m}}^{ - 1}}$)
A. 786 nm
B. 486 nm
C. 586 nm
D. 986 nm

Answer 1

We know that Rydberg's formula is used to calculate the wavelengths of the spectral lines in different series of hydrogen spectrum. Here, the two energy levels and values of Rydberg constant are given. So, we have to use the Rydberg formula, that is, $\dfrac{1}{\lambda } = {R_H}\left( {\dfrac{1}{{{n_1}^2}} - \dfrac{1}{{{n_2}^2}}} \right)$

Complete step by step answer:
The Rydberg formula is,
$\dfrac{1}{\lambda } = {R_H}\left( {\dfrac{1}{{{n_1}^2}} - \dfrac{1}{{{n_2}^2}}} \right)$
…… (1)
Where, $\lambda$ is wavelength, ${R_H}$ is Rydberg constant, ${n_1}$ is quantum number of initial state and ${n_1}$is quantum number of initial state.
Now, come to the question. The value of ${n_1}$ is 2, value of ${n_2}$ is 4 and Rydberg constant is ${R_H} = 109678\,{\rm{c}}{{\rm{m}}^{ - 1}}$.
Now, we have to put all the values in equation (1).
$\dfrac{1}{\lambda } = 109678\,{\rm{c}}{{\rm{m}}^{ - 1}}\left( {\dfrac{1}{{{2^2}}} - \dfrac{1}{{{4^2}}}} \right)$
$\Rightarrow \dfrac{1}{\lambda } = 109678\,{\rm{c}}{{\rm{m}}^{ - 1}}\left( {\dfrac{1}{4} - \dfrac{1}{{16}}} \right)$
$\Rightarrow \dfrac{1}{\lambda } = 109678\,{\rm{c}}{{\rm{m}}^{ - 1}}\left( {\dfrac{{4 - 1}}{{16}}} \right)$
$\Rightarrow \dfrac{1}{\lambda } = 109678\,{\rm{c}}{{\rm{m}}^{ - 1}}\left( {\dfrac{3}{{16}}} \right)$
$\Rightarrow \dfrac{1}{\lambda } = \dfrac{{16}}{{329034}}{\rm{c}}{{\rm{m}}^{ - 1}}$
$\Rightarrow \lambda = 4.86 \times {10^{ - 5}}\,{\rm{cm}}$
Now, we have to convert the value of wavelength to nm. The conversion factor is $\dfrac{{1\,{\rm{nm}}}}{{{{10}^{ - 7}}\,{\rm{cm}}}}$.
$\Rightarrow 4.86 \times {10^{ - 5}}\,{\rm{cm = }}4.86\,{\rm{cm}} \times {10^{ - 5}} \times \dfrac{{1\,{\rm{nm}}}}{{{{10}^{ - 7}}\,{\rm{cm}}}}$
$\Rightarrow 4.86\, \times {10^{ - 5}}{\rm{cm}} = 4.86 \times {10^{ - 2}}\,{\rm{nm}} = {\rm{486}}\,{\rm{nm}}$
Therefore, the value of wavelength is 486 nm. Hence, the correct answer is B, that is, 486 nm.

Additional Information: Let’s discuss the hydrogen spectrum. On passing electric discharge through hydrogen gas enclosed in a discharge tube under low pressure and the emitted light is analysed by a spectroscope, the spectrum consists of a large number of lines which are grouped into different series. The series appear in different regions of light named after their discoveries. The complete spectrum is termed as hydrogen spectrum.

Note: In Rydberg formula, the value of ${n_1}$ is constant while that of ${n_2}$changes. For example,
For the Lyman series, ${n_1}$=1 , ${n_2}$=2,3,4,5……
For the Balmer series, ${n_1}$=2 , ${n_2}$=3,4,5,6……
For the Paschen series,${n_1}$=3 ,${n_2}$=4,5, 6,7……
For the Brackett series, ${n_1}$=4 , ${n_2}$=5,6,7,8…..
For the Lyman series, ${n_1}$=5 , ${n_2}$=6,7,8,9……
Thus, by substituting the values of ${n_1}$and ${n_2}$ in the Rydberg formula , wavelengths of different spectral lines can be calculated. When ${n_1}$=2, the Rydberg expression is termed as Balmer’s formula.