Question

Find the Q-value and the kinetic energy of the emitted α-particle inthe α-decay of (a) 22688 Ra and (b) 22086 Rn .Given m ( 22688 Ra ) = 226.02540 u, m ( 22286 Rn ) = 222.01750 u,m ( 22286 Rn ) = 220.01137 u, m ( 21684 Po ) = 216.00189 u.

Answer 1

Given, the mass of R 88 226 ais 226.02540 u, the mass of R 86 222 a is 226.01137 u, the mass of R 86 222 n is 222.01750 uand mass of P 84 216 o is 216.00189 u.

(a)

The alpha particle decay of R 88 226 a emits helium nucleus.

R 88 226 a  →   R 86 222 a+ H 2 4 e

The Q-value of the reaction is given as,

Q-value=( m i − m f ) c 2 =[ m( R 88 226 a )−{ m( R 86 222 a )+m( H 2 4 e ) } ] c 2

Where, the sum of initial mass is m i and the sum of final mass is m f .

By substituting the given values in the above equation, we get

Q-value=[ 226.02540−{ 222.01750+4.002603 } ] ( c ) 2 u =[ 226.02540−226.020103 ] ( c ) 2 u =0.005297 c 2  u

We know that,

1 u=931.5  MeV/ c 2

Therefore, the Q-value is,

Q-value=0.005297×931.5 =4.94 MeV

The kinetic energy of α particle is given as,

K.E.= A i A f ×Q-value (1)

Where, the mass number before decay is A i and the mass number after decay is A f .

By substituting the corresponding values in the above equation, we get

K.E.= 222 226 ×4.94 =4.85 MeV

Thus, the Q-value and the kinetic energy in alpha decay of R 88 226 a is 4.94 MeV and 4.85 MeV respectively.

(b)

The alpha particle decay of R 88 226 a emits helium nucleus.

R 86 222 n  →   P 84 216 o+ H 2 4 e

The Q-value of the reaction is given as,

Q-value=( m i − m f ) c 2 =[ m( R 88 226 n )−{ m( P 84 216 o )+m( H 2 4 e ) } ] c 2

By substituting the values in the above equation, we get

Q-value=[ 220.01137−{ 216.00189+4.002603 } ] ( c ) 2 u =[ 226.02540−220.004493 ] ( c ) 2 u =6.02097 c 2  u =0.006877 c 2 ×931.5 MeV/ c 2 =6.41 MeV

By substituting the values in equation (1), we get

K.E.= 216 220 ×6.41 =6.29 MeV

Thus, the Q-value and kinetic energy in alpha decay of R 86 222 n is 6.41 MeV and 6.29 MeV respectively.