Login/Sign Up
Home » The Q value of a nuclear reaction A + b → C + d is defined by
The Q value of a nuclear reaction A + b → C + d is defined by
CBSE | Class 12 | NCERT | Nuclei | Physics
The Q value of a nuclear reaction A + b → C + d is defined by

Q = [ mA+ mb− mC− md]c2 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.

  • {}_{1}^{1}H+{}_{1}^{3}H\to {}_{1}^{2}H+{}_{1}^{2}H
  • {}_{6}^{12}C+{}_{6}^{12}C\to {}_{10}^{20}Ne+{}_{2}^{4}He

Atomic masses are given to be

    \[m({}_{1}^{2}H)=2.014102u\]

    \[m({}_{1}^{3}H)=3.016049u\]

m({}_{6}^{12}C)=12.000000u

m({}_{10}^{20}Ne)=19.992439

Ans

(i) The given nuclear reaction is:

{}_{1}^{1}H+{}_{1}^{3}H\to {}_{1}^{2}H+{}_{1}^{2}H

We are given that

    \[m({}_{1}^{1}H)=2.014102u\]

    \[m({}_{1}^{2}H)=2.014102u\]

    \[m({}_{1}^{3}H)=3.016049u\]

So, the Q equation of the reaction is given by –

    \[Q=[m({}_{1}^{1}H)+m({}_{1}^{3}H)-2m({}_{1}^{2}H)]{{c}^{2}}\]

= [ 1.007825 + 3.016049 – 2 x 2.014102] c2

Q = ( – 0.00433 c2) u

We have 1 u = 931.5 MeV/c2

Q = -0.00433 x 931.5 = – 4.0334 MeV

The negative Q-value of the reaction shows that the reaction is endothermic.

(ii)

The given nuclear reaction is:

{}_{6}^{12}C+{}_{6}^{12}C\to {}_{10}^{20}Ne+{}_{2}^{4}He

We are given the values –

m({}_{6}^{12}C)=12.000000u

m({}_{10}^{20}Ne)=19.992439

So, the Q equation of the reaction is given by –

    \[Q=[2m({}_{6}^{12}C)-m({}_{10}^{20}Ne)-m({}_{2}^{4}He)]{{c}^{2}}\]

Q = [ 2 x 12.000000 – 19.992439 – 4.002603 ] c2

Q = [ 0.004958 c2] u

Q = 0.004958 x 931.5 = 4.618377 MeV

Since we obtained positive Q-value, it can be concluded that the reaction is exothermic.

i

More Material