An ionic atom equivalent to a hydrogen atom h | vedclass

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(A) According to the Rydberg formula for hydrogen-like ions,the wavelength $\lambda$ of a spectral line is given by: $\frac{1}{\lambda} = R Z^2 \left(

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$He^+$

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(A) According to the Rydberg formula for hydrogen-like ions,the wavelength $\lambda$ of a spectral line is given by: $\frac{1}{\lambda} = R Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.For a hydrogen atom,$Z = 1$,so $\frac{1}{\lambda_H} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.For the ionic atom,$\frac{1}{\lambda_{ion}} = R Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.Given that $\lambda_{ion} = \frac{1}{4} \lambda_H$,we have $\frac{1}{\lambda_{ion}} = 4 \frac{1}{\lambda_H}$.Substituting the expressions,we get $R Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) = 4 R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.This simplifies to $Z^2 = 4$,which gives $Z = 2$.The element with atomic number $Z = 2$ is Helium $(He)$. Since it is an ion,it is $He^+$. Therefore,the correct option is $A$.

An ionic atom equivalent to a hydrogen atom has a wavelength equal to $1/4$ of the wavelength of the corresponding hydrogen line. The ion is:

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