For a H atom,the wavelength for the highest e | vedclass

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(B) The Rydberg formula for wavelength is given by $\frac{1}{\lambda} = R_H Z^2 (\frac{1}{n_1^2} - \frac{1}{n_2^2})$.For the highest energy transition

Vedclass · Accepted Answer

$22.8$

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(B) The Rydberg formula for wavelength is given by $\frac{1}{\lambda} = R_H Z^2 (\frac{1}{n_1^2} - \frac{1}{n_2^2})$.For the highest energy transition in $H$ atom $(n_1=1, n_2=\infty)$,$\lambda_H = 91.2 \ nm$.For $He^{+}$,$Z=2$. The transition corresponds to the same energy levels $(n_1=1, n_2=\infty)$.Thus,$\lambda_{He^+} = \frac{\lambda_H}{Z^2} = \frac{91.2}{2^2} = \frac{91.2}{4} = 22.8 \ nm$.

For a $H$ atom,the wavelength for the highest energy transition is $91.2 \ nm$. Calculate the corresponding wavelength for $He^{+}$ in $nm$.

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