Modern Physics Ques 22
- Hydrogen $\left({ } _1 H^{1}\right)$, deuterium $\left({ } _1 H^{2}\right)$, singly ionised helium $\left({ } _2 He^{4}\right)^{+}$and doubly ionised lithium $\left({ } _3 Li^{8}\right)^{++}$all have one electron around the nucleus. Consider an electron transition from $n=2$ to $n=1$. If the wavelengths of emitted radiation are $\lambda _1, \lambda _2, \lambda _3$ and $\lambda _4$, respectively for four elements, then approximately which one of the following is correct?
(2014 Main)
(a) $4 \lambda _1=2 \lambda _2=2 \lambda _3=\lambda _4$
(b) $\lambda _1=2 \lambda _2=2 \lambda _3=\lambda _4$
(c) $\lambda _1=\lambda _2=4 \lambda _3=9 \lambda _4$
(d) $\lambda _1=2 \lambda _2=3 \lambda _3=4 \lambda _4$
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Answer:
Correct Answer: 22.(c)
Solution:
Formula:
Wavelength Corresponding To Spectral Lines:
- For hydrogen atom, we get
$\frac{1}{\lambda}=R Z^{2} (\frac{1}{1^{2}}-\frac{1}{2^{2}})$
$\Rightarrow \frac{1}{\lambda _1}=R(1)^{2} (\frac{3}{4}) $
$\Rightarrow \frac{1}{\lambda _2}=R(1)^{2} (\frac{3}{4} ) $
$\Rightarrow \frac{1}{\lambda _3}=R(2)^{2} (\frac{3}{4}) $
$\Rightarrow \frac{1}{\lambda _4}=R(3)^{2} (\frac{3}{4} ) $
$\Rightarrow \frac{1}{\lambda _1}=\frac{1}{4 \lambda _3}=\frac{1}{9 \lambda _4}=\frac{1}{\lambda _2}$
BETA
