Question 473: 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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