Question

Calculate the wavelength and energy of radiation emitted for the electronic transition from infinity to ground state of one of the hydrogen atom.
Given $=3\times {10}^8$ m/sec.
$R=1.09678\times {10}^7{m}^{-1},h=6.6256\times {10}^{-34}$ Js.

• A. $\lambda =9.11\times {10}^{-6},E=217.9\times {10}^{-23}kJ$
• B. $\lambda =8.11\times {10}^{-8},E=317.9\times {10}^{-23}kJ$
• C. $\lambda =9.11\times {10}^{-8},E=217.9\times {10}^{-23}kJ$
• D. none of the above

Answer 1

The correct option is C $\lambda =9.11\times {10}^{-8},E=217.9\times {10}^{-23}kJ$

According to Rydberg equation $\overline{v}=\frac{1}{\lambda }=R{Z}^2[\frac{1}{n_1^2}-\frac{1}{n_2^2}]$
Here $Z=1,R=1.09678\times {10}^7{m}^{-1},n_1=1,n_2=\infty$
$\therefore \frac{1}{\lambda }=1.09678\times {10}^7[\frac{1}{1}-\frac{1}{\infty }]$
$\lambda =\frac{1}{1.09678\times {10}^7}=0.911\times {10}^{-7}=9.11\times {10}^{-8}$ m
energy is given by $E=\frac{hc}{\lambda }=\frac{6.6256\times {10}^{-34}\times 3\times {10}^8}{9.11\times {10}^{-8}}=2.1803\times {10}^{-18}$ Joule $=218.03\times {10}^{-20}J$
or $218.03\times {10}^{-23}kJ$

When we substitute $\frac{1}{\lambda }=1.09678\times {10}^7$ in above expression we will get
$E=217.9\times {10}^{-23}$kJ