The shortest wavelength in the Lyman series i | vedclass

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(A) For the shortest wavelength of the Lyman series,the transition occurs from $n_2 = \infty$ to $n_1 = 1$.Using the Rydberg formula: $\frac{1}{\lambd

Vedclass · Accepted Answer

$121.6$

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(A) For the shortest wavelength of the Lyman series,the transition occurs from $n_2 = \infty$ to $n_1 = 1$.Using the Rydberg formula: $\frac{1}{\lambda_{min}} = R \left( \frac{1}{1^2} - \frac{1}{\infty^2} ight) = R = \frac{1}{91.2 \, nm}$.For the largest wavelength of the Lyman series,the transition occurs from $n_2 = 2$ to $n_1 = 1$.Using the Rydberg formula: $\frac{1}{\lambda_{max}} = R \left( \frac{1}{1^2} - \frac{1}{2^2} ight) = R \left( 1 - \frac{1}{4} ight) = \frac{3}{4} R$.Substituting $R = \frac{1}{91.2 \, nm}$ into the equation:$\frac{1}{\lambda_{max}} = \frac{3}{4} 	imes \frac{1}{91.2 \, nm}$.$\lambda_{max} = \frac{4}{3} 	imes 91.2 \, nm = 121.6 \, nm$.

The shortest wavelength in the Lyman series is $91.2 \, nm$. The largest wavelength of the series is.....$nm$

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