Stationery nucleus `.^(238)U` decays by a emission generaring a total kinetic energy T:
`._(92)^(238) rarr ._(90)^(234)Th +._2^4 alpha`
What is the kinetic energy of the `alpha`-particle?

Options

A. slightly less than `T`
B. `T//2`
C. slightly less than `T`
D. slightly greater than `T`

Answer

C

Explanation

( c) Let then kinetic energy of the `alpha-`particle be `E_alpha` and that of the thorium Th be `E_(Th)`
The ratio of kinetic energies is
`(E_alpha)/(E_(th)) = ((1)/(2) m_alpha v_alpha^2)/((1)/(2) m_(th) v_(th)^2) = ((m_alpha)/(m_(th))) ((v_alpha)/(v_(th)))^2` ....(1)
By conservation of momentum the momentum of `alpha-`particle and that of the recoiling thorium must be equal. Thus,
`m_alpha v_alpha = m_(th) v_(th)`
or `(v_(alpha))/(v_(th)) = (m_(th))/(_(alpha))` ...(2)
Subst. (2) into (1), we have
`(E_(alpha))/(E_(th)) = ((m_alpha)/(m_(th)))((m_(th))/(m_(alpha)))^2 = (m_(th))/(m_(alpha)) = (234)/(4) = 58.5`
Thus, the kinetic energy is the `alpha-`particle expressed as the fraction of the total kinetic energy `T` is given by
`E_alpha = (58.5)/(1 + 58.5) T = (58.5)/(59.5) T = 0.98 T`
which is slightly less than `T`.