In a hydrogen atom, the electron jumps from n=1 level to the n=4 level. What is the wavelength of the emitted or absorbed photon?
the wavelength of the photon is given by lambda=hc/E where h... View the full answer
- also you can get by finding the energy difference that is =n=1=-13.6ev ,n=4=0.85ev ,the difference is 12.75ev convert to joules wihich is multiply by 1.6*10^-19=2.04*10^-18j, using formula of finding lambda= (6.63*10^-34j*3.0*10^8)/(2.04*10^-18j)=9.75*10^-8m
- May 15, 2018 at 11:30am