MCQ
The separation energy of the electron present in the shell $n = 3$ is $1.51\, eV.$ What is the energy in the first exicted state ? ............ $\mathrm{eV}$
- A$-1.51$
- ✓$-3.4$
- C$+1.51$
- D$+3$
✓
Answer
Correct option: B.
$-3.4$
b
1st excited state $=n=2$
1st excited state $=n=2$
Energy $=-136 \times \frac{2}{n^{2}}$$\alpha \frac{1}{n^{2}}$
$\frac{E_{3}}{E_{2}}=\frac{1/(3)^2}{1/(2)^2}$
$\frac{E_{3}}{E_{2}}=\frac{4}{9}$
$E_{2} =\frac{9}{4} \times E_{3}$
$=\frac{9}{4} \times 1.51$
$={-3.4}$
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