Question

The wavelength of radiation emitted, when in $H{e}^{+}$electron falls from infinity to stationary state would be $\left(R=1.097\times 10^7{m}^{-1}\right)$

• A. $2.2\times 10^{-8}m$
• B. $22\times 10^{-9}m$
• C. $120m$
• D. $22\times 10^{7}m$

Answer 1

Answer: (A)

$n_1=1$ For $H{e}^{\oplus }z=2$
$n_2=\alpha$ given
$\frac{1}{{\lambda }_{H{e}^{\oplus }}}=R\left[\frac{1}{1^2}-\frac{1}{{\alpha }^2}\right]\times (2{)}^2$
$\frac{1}{{\lambda }_{H{e}^{\oplus }}}=109678\times 4c{m}^{-1}$
${\lambda }_{H{e}^{\oplus }}=\frac{1}{109678\times 4}=\frac{1}{438712}=2.2\times 10^{-6}cm$
$=2.2\times 10^{-8}m$