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.

 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