Question

Question: An excited hydrogen atom emits light with a wavelength of$\mathbf{397}\mathbf{.}\mathbf{2}\mathbf{}\mathbf{nm}$to reach the energy level for which$\mathbf{n}\mathbf{=}\mathbf{2}$. In which principal quantum level did the electron begin?

Short answer

The principal quantum level for the initial level of an electron to reach the energy level of $\mathrm{n}=2$ the energy state is $7$

Step-by-step solution

Concept used

Electronic transition is a phenomenon that occurs when electrons are emitted or absorbed from one energy level to another. If a transition from a lower to a higher energy level is noticed, absorption occurs. If you go from a greater to a lower energy level, you'll emit something.

Expression for the energy of electrons in a hydrogen atom is as follows:

$\mathrm{E}=-2.178\times {10}^{-18}\mathrm{J}\left(\mathrm{Z}2\mathrm{n}2\right)$

Where,

$\mathrm{E}$is the energy of electrons.

$\mathrm{Z}$is the atomic number.

$\mathrm{n}$is integer.

Expression for the wavelength of the emitted photon

Expression for wavelength of emitted photon is as follows:

$\mathrm{\lambda }=\mathrm{hc}∆\mathrm{E}$ …….. (1)

Where,

$\mathrm{\lambda }$is the wavelength of the emitted photon.

h is Planck's constant.

c is the speed of light.

$∆\mathrm{E}$is energy difference

Rearrange equation (1) for$∆\mathrm{E}$.

$∆\mathrm{E}=\mathrm{hc\lambda }$ …….. (2)

Value of h is $6.626\times {10}^{-34}Js$.

Value of c is$2.9979\times {10}^8\mathrm{m}/\mathrm{s}$.

Value of $\mathrm{\lambda }$ is $397.2\mathrm{nm}$.

Substitute the values in equation (2),

$$\begin{gathered} ∆\mathrm{E}=\mathrm{hc\lambda } \\ =\left(6.626\times {10}^{-34}\mathrm{Js}\right)\left(2.9979\times {10}^8\mathrm{m}/\mathrm{s}\right)\left(397.2\mathrm{nm}\right)\left({10}^9\mathrm{m}1\mathrm{nm}\right) \\ =5.00103\times {10}^{-19}\mathrm{J} \end{gathered}$$

Since an excited hydrogen atom emits light, this energy is taken to be negative.

Expression for energy difference between energy levels of a hydrogen atom is as follows:

$∆\mathrm{E}=-2.178\times {10}^{18}\mathrm{J}\left(1\mathrm{nfinal}2-1\mathrm{ninitial}2\right)$ ……. (3)

Value of n final is 2

Value of $∆\mathrm{E}$is $-5.00103\times {10}^{-19}\mathrm{J}$.

Substitute the values in equation (3),

$$\begin{gathered} ∆\mathrm{E}=-2.178\times {10}^{18}\mathrm{J}\left(1\mathrm{nfinal}2-1\mathrm{ninitial}2\right) \\ -5.00103\times {10}^{-19}\mathrm{J}=-2.178\times {10}^{18}\mathrm{J}\left(1\left(2\right)2-1\mathrm{ninitial}2\right) \end{gathered}$$

The value of n initial is calculated as follows:

n initial

$$\begin{gathered} \mathrm{n}\mathrm{initial}2=49.0 \\ \mathrm{n}\mathrm{initial}=7 \end{gathered}$$

Hence, the principal quantum level for the initial level of an electron to reach the energy level of$\mathrm{n}=2$ the energy state is$7$