Question

The wavelength of radiation emitted, when in $H{e}^{+}$ electron falls from infinity to ground state would be:

$\left(R=1.097\times {10}^7{m}^{-1}\right)$

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

Answer 1

The correct option is A $2.2\times {10}^{-8}m$

Energy of ground state of $H{e}^{+}=13.6\times 4$ $=54.4$ eV

$\therefore \frac{hc}{\lambda }=54.4$ eV

$\Rightarrow \lambda =\frac{6.625\times {10}^{-34}\times 3\times {10}^8}{54.4\times 1.6\times {10}^{-19}}$

$\Rightarrow \lambda =2.2\times {10}^{-8}$m.

Thus, the correct option is $A$