Structure of Atom - Result Question 15
####15. The frequency of radiation emitted when the electron falls from $n=4$ to $n=1$ in a hydrogen atom will be (Given ionization energy of $H=2.18 \times 10^{-18} J^{-1} am^{-1}$ and $h=6.625 \times 10^{-34} J s$ )
[2004]
(a) $1.54 \times 10^{15} s^{-1}$
(b) $1.03 \times 10^{15} s^{-1}$
(c) $3.08 \times 10^{15} s^{-1}$
(d) $2.00 \times 10^{15} s^{-1}$
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Solution:
- (c) I.E. $=E_1-E _{\infty}$
$2.18 \times 10^{-18}=E_1-0$
$E_1=2.18 \times 10^{-18} J^{- \text{atom }^{-1}}$
$\Delta E=E_1(\frac{1}{n_1^{2}}-\frac{1}{n_2^{2}})$
$\Rightarrow h v=2.18 \times 10^{-18}(\frac{1}{1^{2}}-\frac{1}{4^{2}})$
$\Rightarrow 6.625 \times 10^{-34} \times v=2.18 \times 10^{-18} \times \frac{15}{16}$
$\Rightarrow \quad v=\frac{2.18 \times 10^{-18} \times 15}{6.625 \times 10^{-34} \times 16}$
$\Rightarrow v=3.08 \times 10^{15} s^{-1}$
