2 hydrogen atoms undergo head on collision and end up with 0 kinetic energy.Each particle then emits photon of wavelength 1028 angstrom.which transition leads to this wavelength? calculate the velocity of hydrogen atom before collision.mass of hydrogen-(1.67*10to the power minus 27 )kg.

Asked by nishac | 22nd Sep, 2010, 09:35: PM

Expert Answer:

 wavelength = 1028 angstrom in the region of UV rays and thus n1=1 for H atom.
we know the rydberg formula as well,
1/λ=R (1/n21-1/n22)
1/λ=R (1/12-1/n22)
n2=2.88 or 3
and the energy released is due to collision and all the K.E  is released in the form of photon.Thus,
v=48151.3 m/sec
v=4.8*104 m/sec

Answered by  | 23rd Sep, 2010, 08:18: AM

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