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(A) The Rydberg formula for the wavelength of emitted radiation is given by $\frac{1}{\lambda} = R Z^2 \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right

Vedclass · Accepted Answer

$5.099$

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(A) The Rydberg formula for the wavelength of emitted radiation is given by $\frac{1}{\lambda} = R Z^2 \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)$.For a fixed transition ($n_i = 4$ to $n_f = 2$),the wavelength $\lambda$ is inversely proportional to the square of the atomic number $Z$,i.e.,$\lambda \propto \frac{1}{Z^2}$.For hydrogen $(H)$,$Z = 1$,so $\lambda_H = 20.397 \, cm$.For helium ion $(He^+)$,$Z = 2$,so $\lambda_{He^+} = \frac{\lambda_H}{Z^2} = \frac{20.397}{2^2} = \frac{20.397}{4} = 5.099 \, cm$.

The wavelength of the energy emitted when an electron transitions from the fourth orbit to the second orbit in a hydrogen atom is $20.397 \, cm$. The wavelength of the energy for the same transition in $He^+$ is .......... $cm$.

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