If the shortest wavelength of the Lyman serie | vedclass

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(B) For the Lyman series,the shortest wavelength corresponds to the transition from $n_2 = \infty$ to $n_1 = 1$.Using the Rydberg formula: $\frac{1}{x

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$\frac{36x}{5}$

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(B) For the Lyman series,the shortest wavelength corresponds to the transition from $n_2 = \infty$ to $n_1 = 1$.Using the Rydberg formula: $\frac{1}{x} = R 	imes Z^2 	imes (\frac{1}{n_1^2} - \frac{1}{n_2^2})$.For $H$ atom $(Z=1)$: $\frac{1}{x} = R 	imes (\frac{1}{1^2} - \frac{1}{\infty^2}) = R$.So,$R = \frac{1}{x}$.The first excited state of the $H$ atom corresponds to $n = 2$.The total energy $E_n$ is given by $E_n = -13.6 	imes \frac{Z^2}{n^2} 	ext{ eV}$.For $n = 2$,$E_2 = -13.6 	imes \frac{1^2}{2^2} = -3.4 	ext{ eV}$.Note: The question asks for the energy value in terms of $x$. Given the options provided,the question implies a relationship between the wavelength and energy constants. Based on the standard Rydberg constant relation $E = \frac{hc}{\lambda}$,the energy of the first excited state is proportional to the Rydberg constant.

If the shortest wavelength of the Lyman series of the $H$ atom is $x$,then calculate the total energy of the $H$ atom in the first excited state.

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