Modern Physics Ques 54
- 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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Solution:
Formula:
- De-excitation energy of hydrogen electron in transition $n=2$ to $n=1$ is
$$ E=13.6 \times \frac{1}{n _f^{2}}-\frac{1}{n _i^{2}} \quad eV=13.6 \frac{1}{1^{2}}-\frac{1}{2^{2}}=10.2 eV $$
Now, energy levels of helium ion’s $\left(He^{+}\right)$electron are (For helium, $Z=2$ )
$$ \begin{array}{r|r} & n=4-6.04 eV \\ \hline 10.2 eV \uparrow & n=3-13.6 eV \\ \hline & n=2 \\ \hline & \quad-54.4 eV \\ \hline \end{array} $$
So, a photon of energy $10.2 eV$ can cause a transition $n=2$ to $n=4$ in a $He^{+}$ion.
Alternate Solution
For $He^{+}$ion, when in $n=1$ state,
$$
BETA
