Modern Physics Ques 69
- Radiation coming from transitions $n=2$ to $n=1$ of hydrogen atoms fall on $He^{+}$ions in $n=1$ and $n=2$ states. The possible transition of helium ions as they absorb energy from the radiation is
(Main 2019, 8 April I)
(a) $n=2 \rightarrow n=3$
(b) $n=1 \rightarrow n=4$
(c) $n=2 \rightarrow n=5$
(d) $n=2 \rightarrow n=4$
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Answer:
Correct Answer: 69.(d)
Solution:
Formula:
Wavelength Corresponding To Spectral Lines:
- De-excitation energy of hydrogen electron in transition $n=2$ to $n=1$ is
$ E=13.6 \times (\frac{1}{n _f^{2}}-\frac{1}{n _i^{2}}) eV=13.6 (\frac{1}{1^{2}}-\frac{1}{2^{2}})=10.2 $ $eV $
Now, energy levels of helium ion’s $\left(He^{+}\right)$electron are (For helium, $Z=2$ )
So, a photon of energy $10.2$ $ eV$ can cause a transition $n=2$ to $n=4$ in a $He^{+}$ion.
BETA
