Question #138599
Two hydrogen atoms moving at the same speed in opposite directions collide up with zero kinetic energy. Each atom then emits light with a wavelength of 121.6 nm. At what speed were the atoms moving before the collision considering that the mass of a hydrogen atom is equal to that of a proton?
1
Expert's answer
2020-10-19T13:20:37-0400

The kinetic energy, K=12mu2K=\frac{1}{2}mu^{2} of hydrogen was released in form of photon.

Energy of a photon, E=hcλE=\frac{hc}{\lambda}

Therefore 12mu2=hcλ\frac{1}{2}mu^{2} =\frac{hc}{\lambda}

12×1.67×1027×\frac{1}{2} \times1.67\times10^{-27}\times u2=6.626×1034×3×108121.6×109u^{2} = \frac{6.626\times10^{-34}\times3\times10^{8}}{121.6\times10^{-9}}

u=4.43×104ms1u=4.43\times10^{4}ms^{-1}


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