The Q value of a nuclear reaction A + b → C + d is defined by Q = [ mA + mb – mC – md ]c 2 where the masses refer to the respective nuclei. Determine from the given data the Q-value of the following reactions and state whether the reactions are exothermic or endothermic. (i) 1 3 2 2 1 1 1 1 H+ H H+ H → (ii) 12 12 20 4 6 6 10 2 C+ C Ne+ He → Atomic masses are given to be m ( 2 1 H ) = 2.014102 u m ( 3 1 H ) = 3.016049 u m ( 12 6 C ) = 12.000000 u m ( 20 10 Ne ) = 19.992439 u

Given: The atomic masses of H 1 2 is 2.014102 u, H 1 3 is 3.016049 u, C 6 12 is 12.000 u and N 10 20 e is 19.992439 u.

(i)

The given nuclear reaction is,

H 1 1 + H 1 3   →   H 1 2 + H 1 2

The Q-value of the reaction is given as,

Q-value=( m i − m f ) c 2 =[ m( H 1 1 )+m( H 1 3 )−2m( H 1 2 ) ] 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=[ 1.007825−3.016049−2×2.014102 ] c 2 =( −0.00433 c 2 ) u

We know that,

1 u=931.5  MeV/ c 2

Therefore, the required Q-value is,

Q-value=−0.00433×931.5 =−4.0334 MeV

Since, the Q-value is negative therefore, the reaction is endothermic in nature.

Thus, the reaction H 1 1 + H 1 3   →   H 1 2 + H 1 2 is endothermic in nature.

(ii)

The given nuclear reaction is,

C 12 6 + C 12 6   →   N 10 20 e+ H 2 4 e

The Q-value of the reaction is given by,

Q-value=( m i − m f ) c 2 =[ 2m( C 12 6 )−m( N 10 20 e )−m( H 2 4 e ) ] c 2

By substituting the given values in equation (4), we get

Q-value=[ 2×12−19.992439−4.002603 ] c 2 =( 0.004958 c 2 ) u =0.004958 c 2 ×931.5  MeV/c 2 =4.618377 MeV

Since, the Q-value is negative therefore the reaction is exothermic in nature.

Thus, the reaction C 12 6 + C 12 6   →   N 10 20 e+ H 2 4 e is exothermic in nature.