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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Answer:
Correct Answer: 88. $(660$ nm)
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 \hspace{1mm} erg=10^{-7} J\right] \end{aligned} $
Also, $\quad \Delta E=\dfrac{h c}{\lambda}$
$ \begin{aligned} \Rightarrow \quad \lambda & =\dfrac{6.625 \times 10^{-34} \times 3 \times 10^{8}}{3.01 \times 10^{-19}} m \\ & =660 \times 10^{-9} m=660\hspace{1mm} nm \end{aligned} $
BETA
