Question

Energy in the nth Bohr's orbit is given by E = $\frac{-2.179\times {10}^{-18}}{n^2}$ Js
Calculate the frequency and wave number of the radiation emitted when an electron jumps from the third orbit to the second orbit ? (h = 6.62 $\times$ 10${}^{-34}$ Js)

• A. 1.56 $\times$ 10${}^{14}$ s${}^{-1}$
• B. 4.56 $\times$ 10${}^{14}$ s${}^{-1}$
• C. 5.56 $\times$ 10${}^{14}$ s${}^{-1}$
• D. 7.56 $\times$ 10${}^{14}$ s${}^{-1}$

Answer 1

Answer: (B)

4.56 $\times$ 10${}^{14}$ s${}^{-1}$

Energy of Bohr's orbit is given by $E=\frac{B}{n^2}=\frac{2.179\times {10}^{-11}}{n^2}erg.$
For the third orbit, $n=3$
$E_2=\frac{2.179\times {10}^{-11}}{9}=-0.2421\times {10}^{-11}erg.$
For the second orbit, $n=2$
$E_3=\frac{2.179\times {10}^{-11}}{4}=-0.54475\times {10}^{-11}erg.$
$E_3-E_2=(0.54475\times {10}^{-11})-(0.2421\times {10}^{-11})=0.30265\times {10}^{-11}$
When electron jumps from the third to second orbit, a proton is emitted, the energy of which is given by
$hv=E_3-E_2$ where $v$ is the frequency
$v=\frac{E_3-E_2}{h}=\frac{0.30265\times {10}^{-11}}{6.62\times {10}^{-27}}=4.572\times {10}^{14}{sec}^{-1}$