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.

$(1990,3 M)$

Show Answer

Solution:

  1. 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

$$ \begin{aligned} & \Delta E=+4 \times 21.7 \times 10^{-19}\left(\frac{1}{9}-\frac{1}{16}\right) J=4.22 \times 10^{-19} J \ & \text { Also, } \because \quad \Delta E=\frac{h c}{\lambda} \ & \therefore \quad \lambda=\frac{h c}{\Delta E}=\frac{6.625 \times 10^{-34} \times 3 \times 10^{8}}{4.22 \times 10^{-19}} m \ &=471 \times 10^{-9} m=471 nm \end{aligned} $$



NCERT Chapter Video Solution

Dual Pane