Atomic Structure - Result Question 112
####88. The energy of the electron in the second and third Bohr’s orbits of the hydrogen atom is $-5.42 \times 10^{-12} erg$ and $-2.41 \times 10^{-12}$ erg respectively. Calculate the wavelength of the emitted light when the electron drops from the third to the second orbit.
(1981, 3M)
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Solution:
- Transition energy $=[-2.41-(-5.42)] \times 10^{-12} erg$
$$ \begin{aligned} & =3.01 \times 10^{-12} erg \ & =3.01 \times 10^{-19} J \quad\left[\because 1 erg=10^{-7} J\right] \end{aligned} $$
Also, $\quad \Delta E=\frac{h c}{\lambda}$
$$ \begin{aligned} \Rightarrow \quad \lambda & =\frac{6.625 \times 10^{-34} \times 3 \times 10^{8}}{3.01 \times 10^{-19}} m \ & =660 \times 10^{-9} m=660 nm \end{aligned} $$
