Question

Calculate the longest and shortest wavelengths of light emitted by electrons in the hydrogen atom that begin in the n=6state and then fall to states with smaller values of n.

Short answer

The longest wavelength of the emitted light is $7462.09\mathrm{nm}$

The shortest wavelength of the emitted light is$93.80\mathrm{nm}$

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,

E is the energy of electrons.

Z is the atomic number.

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.

Calculate the Longest wavelength of emitted light

When an electron jumps from the sixth to fifth energy level, its energy difference is the least as compared to other levels. Since wavelength is inversely related to energy difference, the wavelength of light emitted by a jump of an electron from the sixth to fifth energy level is the longest.

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

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

The value of n final is 5.

The value of n initial is 6.

Substitute the values in equation (2),

$$\begin{gathered} ∆\mathrm{E}=-2.178\times {10}^{-18}\mathrm{J}(1\mathrm{nfinal}2-1\mathrm{ninital}2) \\ =-2.178\times {10}^{-18}\mathrm{J}\left(1\right(5)2-1(6\left)2\right) \\ =-0.02662\times {10}^{-18}\mathrm{J} \end{gathered}$$

The negative sign indicates that energy is emitted when an electron jumps from the sixth to fifth energy level.

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

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

Value of is $∆\mathrm{E}\mathrm{is}0.02662\times {10}^{-18}\mathrm{J}$

Substitute the values in equation (1),

$$\begin{gathered} \mathrm{\lambda }=\mathrm{hc}∆\mathrm{E} \\ =(6.626\times {10}^{-34}\mathrm{J}-\mathrm{s})(2.9979\times {10}^8\mathrm{m}/\mathrm{s})0.02662\times {10}^{-18}\mathrm{J} \\ (7.46209\times {10}^3\mathrm{m})\left(1\mathrm{nm}{10}^{-9}\mathrm{m}\right) \\ =7462.09\mathrm{nm} \end{gathered}$$

Hence, the longest wavelength of the emitted light is $7462.09\mathrm{nm}$.

Calculate the Shortest wavelength of emitted light

When an electron jumps from sixth to first energy level, its energy difference is the highest as compared to other levels. Since wavelength is inversely related to energy difference, the wavelength of light emitted by a jump of an electron from the sixth to the first energy level is the shortest.

Value of n final is 1.

The value of n initial is 6.

Substitute the values in equation (2),

$$\begin{gathered} ∆\mathrm{E}=-2.178\times {10}^{-18}\mathrm{J}(1\mathrm{nfinal}2-1\mathrm{ninital}2) \\ =-2.178\times {10}^{-18}\mathrm{J}\left(1\right(1)2-1(6\left)2\right) \\ =-2.1175\times {10}^{-18}\mathrm{J} \end{gathered}$$

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

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

Value of $∆\mathrm{E}\mathrm{is}2.1175\times {10}^{-18}\mathrm{J}$

Substitute the values in equation (1),

$$\begin{gathered} \mathrm{\lambda }=\mathrm{hc}∆\mathrm{E} \\ =(6.626\times {10}^{-34}\mathrm{J}-\mathrm{s})(2.9979\times {10}^8\mathrm{m}/\mathrm{s})2.1175\times {10}^{-18}\mathrm{J} \\ (9.380\times {10}^{-8}\mathrm{m})\left(1\mathrm{nm}{10}^9\mathrm{m}\right) \\ =93.80\mathrm{nm} \end{gathered}$$

Hence, the shortest wavelength of the emitted light is $93.80\mathrm{nm}$.