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 :
$E_n=\dfrac{-21.7 \times 10^{-19}}{n^{2}} J$
Calculate the longest wavelength of electron from the third Bohr’s orbit of the $He^{+}$ion.
(1990,3 M)
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Answer:
Correct Answer: 83. $(471$ nm)
Solution:
- For H-like species, the energy of stationary orbit is expressed as
$E(X)=Z^{2} \times E(H)$
$\Rightarrow$ For $He^{+}(Z=2)$
$E=-\dfrac{4 \times 21.7 \times 10^{-19}}{n^{2}} J$
For longest wavelength transition from 3rd orbit, electron must jump to 4th orbit and the transition energy can be determined as
$\begin{aligned} & \Delta E=+4 \times 21.7 \times 10^{-19}\left(\dfrac{1}{9}-\dfrac{1}{16}\right) J=4.22 \times 10^{-19} J \\ & \text { Also, } \because \quad \Delta E=\dfrac{h c}{\lambda} \\ & \therefore \quad \lambda=\dfrac{h c}{\Delta E}=\dfrac{6.625 \times 10^{-34} \times 3 \times 10^{8}}{4.22 \times 10^{-19}} m \\ &=471 \times 10^{-9} m=471 \hspace{1mm} nm \end{aligned}$
BETA
