Question

In H-spectrum, longest wavelength of Lyman is $120$ nm and shortest wavelength of Balm From this data, find the longest wavelength of the photon, that can ionize this H-atom
A proton and an electron, both at rest initially, combine to form an H-atom in a ground state. A is emitted in this process. Find the wavelength (in nm ) of this photon.

Answer 1

In $H$ spectrum longest wavelength if lyman$=120nm$

$\frac{1}{\lambda }=R_H(\frac{1}{1^2}-\frac{1}{{\infty }^2})$

$\lambda =\frac{1}{R_H}\frac{1}{R_H}=120nm$

Shortest wavelength of Balmer Series

$\frac{1}{\lambda }=R_H(\frac{1}{2^2}-\frac{1}{3^2})$

$\frac{1}{\lambda }=R_H(\frac{1}{4}-\frac{1}{9})$

$\lambda =\frac{36}{5R_H}$

$\frac{36}{5}x120$

$=864nm$

Energy emitted while forming a $Hatom$

$=2.176\times {10}^{-18}joule$

$\frac{46}{\lambda }=2.176\times {10}^{-18}$

$\lambda =\frac{46}{2.176\times {10}^{-18}}$

$=\frac{6.626\times {10}^{-34}\times 3\times {10}^8}{2.176\times {10}^{-18}}$

$=\frac{19.878}{2.176}\times {10}^{-8}$

$=9.135\times {10}^{8}m$

$=913.5A^o$