An electron of energy 20 eV collides with a hydrogen atom in the ground state. As a result of the collision, the atom is excited to a higher energy state and the electron is scattered with reduced velocity. The atom subsequently returns to its ground state with emission of radiation of wavelength 1.216 × 10 –7 m. Find the velocity of the scattered electron.
Text Solution
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Sol. Energy lost by the electron in exciting the hydrogen atom equals the energy corresponding to λ = 1.216 × 10 –7 m
E = 
= 
= 16.36 × 10 –19 J.
Now the initial KE of electron is 20 eV = 32 × 10 –19 J, Hence the kinetic energy of the scattered electron is 32 × 10 –19 – 16.36 × 10 –19 = 15.64 × 10 –19 J.
Velocity V of the scattered electron is given by 
mv 2 = E
V = 
= 
= 1.86 × 10 6 ms –1 .
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