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Q.
The wavelength of radiation emitted, when in $He ^{+}$electron falls from infinity to stationary state would be $\left(R=1.097 \times 10^{7} m ^{-1}\right)$
Structure of Atom
Solution:
$n_{1}=1 $ For $He^{\oplus} z=2$
$n_{2}=\alpha$ given
$\frac{1}{\lambda_{ He ^{\oplus}}}=R\left[\frac{1}{1^{2}}-\frac{1}{\alpha^{2}}\right] \times(2)^{2}$
$\frac{1}{\lambda_{ He ^{\oplus}}}=109678 \times 4\, cm ^{-1}$
$\lambda_{ He ^{\oplus}}=\frac{1}{109678 \times 4}=\frac{1}{438712}=2.2 \times 10^{-6} cm$
$=2.2 \times 10^{-8} m$