Atomic Structure - Result Question 107
####83. According to Bohr’s theory, the electronic energy of hydrogen atom in the $n$th Bohr’s orbit is given by :
$$ E _n=\frac{-21.7 \times 10^{-19}}{n^{2}} J $$
Calculate the longest wavelength of electron from the third Bohr’s orbit of the $He^{+}$ion.
$(1990,3 M)$
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Solution:
- For H-like species, the energy of stationary orbit is expressed as $E(X)=Z^{2} \times E(H)$
$\Rightarrow$ For $He^{+}(Z=2)$
$$ E=-\frac{4 \times 21.7 \times 10^{-19}}{n^{2}} J $$
For longest wavelength transition from 3rd orbit, electron must jump to 4 th orbit and the transition energy can be determined as
$$ \begin{aligned} & \Delta E=+4 \times 21.7 \times 10^{-19}\left(\frac{1}{9}-\frac{1}{16}\right) J=4.22 \times 10^{-19} J \ & \text { Also, } \because \quad \Delta E=\frac{h c}{\lambda} \ & \therefore \quad \lambda=\frac{h c}{\Delta E}=\frac{6.625 \times 10^{-34} \times 3 \times 10^{8}}{4.22 \times 10^{-19}} m \ &=471 \times 10^{-9} m=471 nm \end{aligned} $$
