Question

the wavelength of the radiation emitted , when in a hydrogen atom electron falls from infinity to stationary state 1 , would be

Answer 1

E= -13.6(1/n^2 - 1/m^2)
hc/W= -13.6(1/n^2 - 1/m^2). {W means wavelength}
hc/ -13.6(1/n^2 - 1/m^2) =W
solve it by substituting h=6.6×10^-34 ,c=3×10^8
n=1 ,m=infinity
(1/infinity=0)

Answer 2

The emitted radiation's wavelength, in hydrogen atom, when electron falls to stationary state 1 from infinity, would be 91 nm.

Explanation:

$\frac{1}{\lambda}=\overline{\mathrm{V}}_{\mathrm{H}}=\overline{\mathrm{R}_{\mathrm{H}}}\left[\frac{1}{\mathrm{n}_{1}^{2}}-\frac{1}{\mathrm{n}_{2}^{2}}\right]$

$=1.097 \times 10^{7}\left[\frac{1}{1^{2}}-\frac{1}{\infty}\right]$

$\therefore \lambda=\frac{1}{1.097 \times 10^{7}} \mathrm{m}$

$\lambda=9.11 \times 10^{-8} \mathrm{m}$

$=91.1 \times 10^{-9} \mathrm{m}$

$\lambda=91.1 \mathrm{nm}\left(1 \mathrm{nm}=10^{-9} \mathrm{m}\right)$