Question 430: Atoms & Nuclei

Question

An excited hydrogen atom emits a photon of wavelength $\lambda$ in returning to ground state. The quantum number $n$ of the excited state is ($R=$ Rydberg's constant)

Accepted Answer

The wave number of emitted photon when electron jumps from $n_1$ to $n_2$ is given by
$v=\frac{1}{\lambda}=R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}ight)$ .... (i)
Here, $n_1=1, n_2=n$
Put the value of $n_1$ and $n_2$ in Eq. (i), we get
$\begin{array}{ll}
\Rightarrow & \frac{1}{\lambda}=R\left[\frac{1}{1^2}-\frac{1}{n^2}ight] \
\Rightarrow & \frac{1}{\lambda R}=\left[1-\frac{1}{n^2}ight] \Rightarrow \frac{1}{n^2}=1-\frac{1}{\lambda R} \
\Rightarrow & \frac{1}{n^2}=\frac{(\lambda R-1)}{\lambda R} \Rightarrow n^2=\frac{\lambda R}{(\lambda R-1)} \
\Rightarrow & n=\sqrt{\frac{\lambda R}{(\lambda R-1)}}
\end{array}$

Answer

$\sqrt{\lambda R(\lambda R-1)}$

Answer

$\sqrt{\frac{\lambda R}{(\lambda R-1)}}$

Answer

$\sqrt{\frac{(\lambda R-1)}{\lambda R}}$

Answer

$\sqrt{\frac{1}{\lambda R(\lambda R-1)}}$

Atoms & Nuclei

An excited hydrogen atom emits a photon of wavelength \lambda in returning to ground state. The quantum number n of the excited state is (R= Rydberg's constant)

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