Question

The wavelength of radiation emitted when an electron in a hydrogen atom makes a transition from an energy level with $n=3$ to a level with $n=2$ is: [Given that $E_n=\frac{-1312}{n^2}kJmo{l}^{-1}$]

• A. $6.54\times {10}^{-7}$ m
• B. $65.6$ nm
• C. $65.6\times {10}^{-7}$ m
• D. any of the above

Answer 1

The correct option is A $6.54\times {10}^{-7}$ m

$\epsilon =hv=\frac{hc}{\lambda }$

$-1312(\frac{1}{3^2}-\frac{1}{2^2})\times {10}^3=\frac{6.626\times {10}^{-34}\times 3\times {10}^8}{\lambda }$

$\lambda =1.091\times {10}^{-30}m.mole$

$1mole=6.023\times {10}^{23}{e}^{-}$ of hydrogen

$\lambda =1.091\times {10}^{-30}\times 6.023\times {10}^{23}$

$\lambda =6.570\times {10}^{-7}m$