Question

Please help!!

Consider the atomic spectra of hydrogen and deuterium. For atomic hydrogen, the first tran- sition in the Balmer series (n = 3 + 2) occurs at 656.45377 nm. At what wavelength would you expect the analogous transition to be observed for atomic deuterium? Would you be able to distinguish these two transitions with a spectrograph that has 0.01 nm resolution?

Answer 1

The wavelengthb of any transtion is related to the energies of the states between which transtion occurs. For hydrogen and deuterium, the energy of the states  will be different as energy of a electron-nucleus system depends upon the reduced mass of the atom which we know is different for hydrogen and Deuterium.

ne {1,2,3,..} where, - Eground = ? 27² where M is the reduced mass; M = memn metmN

So,

E transition energy Acties in For hydrogen and proton, The on the imamo Mo - ma me (mot me me mp (or, as . do AE = he 1 lt me md Tlt me

So , m_{e = 0.00054858} m_d= 2.014102 amu m_p=   1.00727647amu

So, $\lambda$_D =  $\lambda$_H( 0.999727938)

= 656.45377X 0.999727938 = 656.275174 nm

The above two lines can be distinguished by using a spectrograph having resolution 0.1nm