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A hydrogen atom is in an excited state of principal quantum number $(n)$, it emits a photon of wavelength $(\lambda)$, when it returns to the ground state. The value of $n$ is
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Verified Answer
The correct answer is:
$\sqrt{\frac{\lambda R}{\lambda R-1}}$
As $\frac{1}{\lambda}=R\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right)$ $\therefore \quad \frac{1}{\lambda}=R\left(\frac{1}{1^{2}}-\frac{1}{n^{2}}\right)$
Multiply both sides by $\lambda$
$1=\lambda R\left(1-\frac{1}{n^{2}}\right)$
or $\frac{1}{\lambda R}=1-\frac{1}{n^{2}}$
or $\quad \frac{1}{n^{2}}=1-\frac{1}{\lambda R}=\frac{\lambda R-1}{\lambda R}$
or $\quad n=\sqrt{\frac{\lambda R}{\lambda R-1}}$
Multiply both sides by $\lambda$
$1=\lambda R\left(1-\frac{1}{n^{2}}\right)$
or $\frac{1}{\lambda R}=1-\frac{1}{n^{2}}$
or $\quad \frac{1}{n^{2}}=1-\frac{1}{\lambda R}=\frac{\lambda R-1}{\lambda R}$
or $\quad n=\sqrt{\frac{\lambda R}{\lambda R-1}}$
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