Question

Question: One of the emission spectral lines for${\mathbf{Be}}^{\mathbf{3}\mathbf{+}}$has a wavelength of$\mathbf{253}\mathbf{.}\mathbf{4}\mathbf{}\mathbf{nm}$for an electronic transition that begins in the state with$\mathbf{n}\mathbf{=}\mathbf{5}$. What is the principal quantum number of the lower-energy state corresponding to this emission?

Short answer

The principal quantum number of lower-energy state that corresponds to emission from $\mathrm{n}=5$is$4$

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 energy of electrons in hydrogen atom is as follows:

$\mathrm{E}=-2.178\times {10}^{-18}\mathrm{J}\left(\frac{{\mathrm{z}}^2}{{\mathrm{n}}^2}\right)$ ……. (1)

Where,

E is the energy of electrons.

Z is the atomic number.

n is an integer.

Calculate the Principal quantum number of lower-energy state

Expression for the wavelength of the emitted photon is as follows:

$\mathrm{\lambda }=\frac{\mathrm{hc}}{∆\mathrm{E}}$

Rearrange above equation for$∆\mathrm{E}$ .

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

Value of h is$6.626\times {10}^{-34}\mathrm{J}.\mathrm{s}$.

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

Value of$\mathrm{\lambda }$is$253.4\mathrm{nm}$.

Substitute the values in equation (2),

$$\begin{gathered} ∆\mathrm{E}=\frac{\mathrm{hc}}{\mathrm{\lambda }} \\ =\frac{\left(6.626\times {10}^{-34}\mathrm{J}.\mathrm{s}\right)\left(2.9979\times {10}^8\mathrm{m}/\mathrm{s}\right)}{\left(253.4\mathrm{nm}\right)\left({\displaystyle \frac{{10}^{-9}\mathrm{m}}{1\mathrm{mm}}}\right)} \\ =7.83902\times {10}^{-19}\mathrm{J} \end{gathered}$$

Since excited electrons in the fifth energy level will come to lower energy levels, it is accompanied by emission of energy and therefore this energy is taken to be negative.

Expression for energy difference between energy levels of is:

$∆\mathrm{E}=\left(-2.178\times {10}^{-18}\mathrm{J}\right)\left({\mathrm{Z}}^2\right)\left(\frac{1}{{\mathrm{n}}_{\mathrm{final}}^2}-\frac{1}{{\mathrm{n}}_{\mathrm{initial}}^2}\right)$ ……. (3)

Value of $\mathrm{Z}$is$4$

Value of$∆\mathrm{E}$is$-7.83902\times {10}^{-19}\mathrm{J}$.

Value of${\mathrm{n}}_{\mathrm{initial}}$is$5$

Substitute the values in equation (3),

$$\begin{gathered} ∆\mathrm{E}=\left(-2.178\times {10}^{-18}\mathrm{J}\right)\left({\mathrm{Z}}^2\right)\left(\frac{1}{{\mathrm{n}}_{\mathrm{final}}^2}-\frac{1}{{\mathrm{n}}_{\mathrm{initial}}^2}\right) \\ -7.83902\times {10}^{-19}\mathrm{J}=\left(-2.178\times {10}^{-18}\mathrm{J}\right)\left(4^2\right)\left(\frac{1}{{\mathrm{n}}_{\mathrm{final}}^2}-\frac{1}{5^2}\right) \\ \left(\frac{1}{{\mathrm{n}}_{\mathrm{final}}^2}-\frac{1}{25}\right)=0.0225 \\ \frac{1}{{\mathrm{n}}_{\mathrm{final}}^2}=0.0225+\frac{1}{25} \\ \frac{1}{{\mathrm{n}}_{\mathrm{final}}^2}=0.0625 \\ {\mathrm{n}}_{\mathrm{final}}^2=16 \\ {\mathrm{n}}_{\mathrm{final}}=\sqrt{16} \\ =4 \end{gathered}$$

Hence, the Principal quantum number of lower-energy state that corresponds to emission from $\mathrm{n}=5$is$4$