Question

An excited hydrogen atom emits a photon of wavelength $\lambda $ in returning to the ground state. If R is the Rydberg constant, then the quantum number n of the excited state will be:
A. $\sqrt{\lambda \text{R}}$
B. $\sqrt{\lambda \text{R-1}}$
C. $\sqrt{\dfrac{\lambda \text{R}}{\lambda \text{R}-1}}$
D. $\sqrt{\lambda \text{R(}\lambda \text{R}-1)}$

Answer 1

Hint: For this problem, we have to use the Rydberg equation that is $\dfrac{\text{1}}{\lambda }\text{ = R}\left( \dfrac{1}{\text{n}_{1}^{2}}-\dfrac{1}{\text{n}_{2}^{2}} \right)$, here $\lambda $is the wavelength, R is known as Rydberg constant and ${{\text{n}}_{1}}\text{ and }{{\text{n}}_{2}}$ are the position of an electron at the ground state and excited state respectively.

Complete step by step solution:
- In the given question, we have to calculate the quantum number which is given as 'n' at the excited state.
- Here, the spectral lines are considered as the dark or light coloured lines which are formed when the hydrogen atom absorbs energy and excites to the excited level from the ground level.
- As a result of excitation the electrons emit the light of different colours.
- Now, in the question, it is given that the wavelength of the photon is $\lambda $, whereas the value of the Rydberg constant is fixed that is R.
- Now, by using the formula of Rydberg equation we will calculate the value of ${{\text{n}}_{2}}$ whose value is given as n.
- The Rydberg equation is:
$\dfrac{\text{1}}{\lambda }\text{ = R}\left( \dfrac{1}{\text{n}_{1}^{2}}-\dfrac{1}{\text{n}_{2}^{2}} \right)$
$\dfrac{\text{1}}{\lambda \text{R}}\text{ = }\dfrac{{{\text{n}}^{2}}-1}{{{\text{n}}^{2}}}$ …. (A)
- The equation (A) can also be written as:
$\lambda \text{R(}{{\text{n}}^{2}}-1\text{) = }{{\text{n}}^{2}}$
$\lambda \text{R}{{\text{n}}^{2}}-{{\text{n}}^{2}}\text{ = }\lambda \text{R}$
${{\text{n}}^{2}}(\lambda \text{R}-1)\text{ = }\lambda \text{R}$
${{\text{n}}^{2}}\text{ = }\dfrac{\lambda \text{R}}{(\lambda \text{R}-1)\text{ }}$
$\text{n = }\sqrt{\dfrac{\lambda \text{R}}{\lambda \text{R}-1}}$
- So, we can see that the quantum number, n is equal to the under the root of the ratio of the $\lambda \text{R}$ and $\lambda \text{R}-1$.

Therefore, option C is the correct answer.

Note: According to the lines emitted by the hydrogen, there are five types of series that are named after the scientist who discovered them that is Lyman series (n = 1), Balmer series (n = 2), Paschen series (n = 3), Brackett series (n = 4) and Pfund series (n = 5).