Question

A hydrogen like atom is in its $5^{th}$ excited state. This excited atom can make transition to the $1^{st}$ excited state by successively emitting two photons of energy $10.2eV$ and $17eV$ respectively. Which of the following option(s) is/are correct?

• A. This hydrogen like atom is Lithium.
• B. For $10.2eV$, photon transition is from $n=6$ to $n=3$.
• C. For $17eV$, photon transition is from $n=4$ to $n=2$.
• D. For $10.2eV$, photon transition is from $n=2$ to $n=1$.

Answer 1

Answer: (A), (B)

The correct options are
A This hydrogen like atom is Lithium.
B For $10.2eV$, photon transition is from $n=6$ to $n=3$.

The total energy emitted when the electron makes transition from $5^{th}$ excited state to $1^{th}$ excited state is given by
$13.6eV{Z}^2(\frac{1}{4}-\frac{1}{36})=10.2eV+17eV$
$13.6Z^2\frac{(9-1)}{36}=27.2$

$\Rightarrow \frac{13.5\times 8}{36}{Z}^2=27.2$

$Z=\sqrt{\frac{272\times 36}{136\times 8}}\Rightarrow Z=3$

$13.6\times 9(\frac{1}{n^2}-\frac{1}{36})=10.2$

By hit and trial we get, for $n=3$ equation is satisfied.
So for $10.2eV$ photon, transition is made from $n=6$ to $n=3$.
and for the photon emission of $17eV$, transition is from $n=3$ to $n=2$.