If the hydrogen atom emits a photon of wavelength $\lambda=4860 \AA$, then the change in the K.E. of

Question

If the hydrogen atom emits a photon of wavelength $\lambda=4860 \AA$, then the change in the K.E. of the electron in the atom is\

Professor PC Thomas Classes · Answer

$\Delta \mathrm{E}=\frac{\mathrm{hc}}{\lambda}=\frac{6.6 	imes 10^{-34} 	imes 3 	imes 10^{8}}{4860 	imes 10^{-10}}$\ $=\quad \mathrm{J}$\ $1 \mathrm{eV}=1.6 	imes 10^{-19} \mathrm{~J}$\ Get the answer is eV .\