Question
An electron in the hydrogen atom initially in the fourth excited state makes a transition to $\mathrm{n}^{\text {th }}$ energy state by emitting a photon of energy 2.86 eV . The integer value of $n$ will be __________ .
✓
Answer
2
$E=13.6\left(\frac{1}{\mathrm{n}_{1}^{2}}-\frac{1}{\mathrm{n}_{1}^{2}}\right)$
$2.86=13.6\left(\frac{1}{\mathrm{n}^{2}}-\frac{1}{5^{2}}\right)$
$\frac{1}{\mathrm{n}^{2}}=0.21+\frac{1}{2.5}$
$\mathrm{n}^{2}=4$
$\mathrm{n}=2$
$E=13.6\left(\frac{1}{\mathrm{n}_{1}^{2}}-\frac{1}{\mathrm{n}_{1}^{2}}\right)$
$2.86=13.6\left(\frac{1}{\mathrm{n}^{2}}-\frac{1}{5^{2}}\right)$
$\frac{1}{\mathrm{n}^{2}}=0.21+\frac{1}{2.5}$
$\mathrm{n}^{2}=4$
$\mathrm{n}=2$
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