Question 524: Atoms & Nuclei

Question

In hydrogen emission spectrum, for any series, the principal quantum number is $n$. Corresponding maximum wavelength $\lambda$ is ( $R=$ Rydberg's constant)

Accepted Answer

In hydrogen emission spectrum, wavelength is given by
$$\frac{1}{\lambda}=R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}ight)$$
For maximum wavelength of any principal quantum number $n$,
$$\begin{aligned}
& n_1=n 	ext { and } n_2=n+1 \
	herefore\quad & \frac{1}{\lambda_{\max }}=R\left[\frac{1}{n^2}-\frac{1}{(n+1)^2}ight] \
& =R\left[\frac{(n+1)^2-n^2}{n^2(n+1)^2}ight] \
& =R \frac{\left(n^2+2 n+1-n^2ight)}{n^2(n+1)^2} \
& \frac{1}{\lambda_{\max }}=\frac{R^2(2 n+1)}{n^2(n+1)^2} \
& 	herefore \quad \lambda_{\max }=\frac{n^2(n+1)^2}{R(2 n+1)}
\end{aligned}$$

Answer

$\frac{R(2 n+1)}{n^2(n+1)}$

Answer

$\frac{n^2(n+1)^2}{R(2 n+1)}$

Answer

$\frac{n^2(n+1)}{R(2 n+1)}$

Answer

$\frac{R(2 n+1)}{n^2(n+1)^2}$

Atoms & Nuclei

In hydrogen emission spectrum, for any series, the principal quantum number is n. Corresponding maximum wavelength \lambda is ( R= Rydberg's constant)

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