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)
 
1/(1028*10-10)=1.097*107(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,
 
1/2mv2=hc/λ
 
(1/2)*1.67*10-27*v2=6.634*10-34*3*108/(1028*10-10)
 
v=48151.3 m/sec
or 
v=4.8*104 m/sec

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

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