Question

One of the visible lines in the hydrogen emission spectrum corresponds to the$n=6ton=2$electronic transmission. What colour light is this transition? See exercise$13.5$.

Short answer

the wavelength of the radiation released is $4.1\times {10}^{-5}cm$so it belongs to the violet region.

Step-by-step solution

Step-by-step-solution

(Identifying the equation to calculate the energy of an electron is given the energy level $\left(n\right)$of the given atom with atomic number Z)

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

Atomic no of hydrogen is $1$

Energy,
role="math" localid="1656398556376" $E=-2.178\times {10}^{-18}J\left[\frac{1^2}{n^2}\right]$

Energy,
$E=2.178\times {10}^{-18}J\left[\frac{1}{n^2}\right]$

(Energy of electron at E2and E6 calculating energy released which is equal to the energy levels)

Energy, $E_6=-2.178\times {10}^{-18}J\left[\frac{1}{6^2}\right]$

Energy,$$\begin{gathered} E_2=-2.178\times {10}^{-18}J\left[\frac{1}{2^2}\right] \\ AE=E_6-E_2 \\ =\left[-2.178\times {10}^{-18}J\left(\frac{1}{6^2}\right)\right]-\left[-2.178\times {10}^{-18}J\left(\frac{1}{2^2}\right)\right] \\ =-2.178\times {10}^{-18}J\left(\frac{1}{6^2}-\frac{1}{2^2}\right) \\ ∆E=0.484\times {10}^{-18}J \end{gathered}$$

(Calculating the wavelength of the radiation released)

$$\begin{gathered} E=\frac{hc}{\lambda } \\ \lambda =\frac{hc}{E} \end{gathered}$$

The velocity of light is $3\times {10}^{8}m/s$

Value of plank’s constant $=6.626\times {10}^{-34}Js$

Wavelength$$\begin{gathered} =\left(6.626\times {10}^{-34}Js\right)\times \frac{3\times {10}^{8}m{s}^{-1}}{0.484\times {10}^{-18}J} \\ =41.07\times {10}^{-34+8-18}m \\ =41.07\times {10}^{-8}m\times \frac{100cm}{1m} \\ \lambda =4.1\times {10}^{-5} \end{gathered}$$