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Ncert Solutions Physics Class 12th Physics Ncert Solutions Class 12th Physics Class 12th Nuclei

13.15 The Q value of a nuclear reaction A + b $\to$ C + d is defined by

Q = [ $m_{A} + m_{b}$ – $m_{C}$ – $m_{d}$ ] $c^{2}$

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

(ii) ${}_{6}^{12}C$ + ${}_{6}^{12}C$ $\to$ ${}_{10}^{20}{N e}$ + ${}_{2}^{4}{H e}$

Atomic masses are given to be

m ( ${}_{1}^{2}H$ ) = 2.014102 u

m ( ${}_{1}^{3}H$ ) = 3.016049 u

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

m ( ${}_{10}^{20}{N e}$ ) = 19.992439 u

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Payal Gupta | Contributor-Level 10

3 months ago

13.15 The given nuclear reaction is

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

Atomic mass

m ( ${}_{1}^{1}H$ ) = 1.007825 u

m ( ${}_{1}^{2}H$ ) = 2.014102 u

m ( ${}_{1}^{3}H$ ) = 3.016049 u

The Q-value of the reaction can be written as:

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

= $[1.007825 + 3.016049 - 2 \times 2.014102] c^{2}$

= (-4.33 $\times 10^{- 3}$ ) $c^{2}$

But 1 u = 931.5 MeV/ $c^{2}$

Q = -4.0334 MeV

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

The given nuclear reaction is

${}_{6}^{12}C$ + ${}_{6}^{12}C$ $\to$ ${}_{10}^{20}{N e}$ + ${}_{2}^{4}{H e}$

Atomic mass

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

m ( ${}_{10}^{20}{N e}$ ) = 19.992439 u

m ( ${}_{2}^{4}{H e}$ ) = 4.002603 u

The Q-value of this reaction is given as:

Q = $[2 m ({}_{6}^{12}C) - m ({}_{10}^{20}{N e}) - m ({}_{2}^{4}{H e})] c^{2}$

= $[2 \times 12.0 - 19.992439 - 4.002603] c^{2}$

=4.958 $\times 10^{- 3} c^{2}$ u

=4.958 $\times 10^{- 3} \times 931.5$

=4.6183 MeV

...more

13.15 The given nuclear reaction is

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

Atomic mass

m ( ${}_{1}^{1}H$ ) = 1.007825 u

m ( ${}_{1}^{2}H$ ) = 2.014102 u

m ( ${}_{1}^{3}H$ ) = 3.016049 u

The Q-value of the reaction can be written as:

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

= $[1.007825 + 3.016049 - 2 \times 2.014102] c^{2}$

= (-4.33 $\times 10^{- 3}$ ) $c^{2}$

But 1 u = 931.5 MeV/ $c^{2}$

Q = -4.0334 MeV

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

The given nuclear reaction is

${}_{6}^{12}C$ + ${}_{6}^{12}C$ $\to$ ${}_{10}^{20}{N e}$ + ${}_{2}^{4}{H e}$

Atomic mass

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

m ( ${}_{10}^{20}{N e}$ ) = 19.992439 u

m ( ${}_{2}^{4}{H e}$ ) = 4.002603 u

The Q-value of this reaction is given as:

Q = $[2 m ({}_{6}^{12}C) - m ({}_{10}^{20}{N e}) - m ({}_{2}^{4}{H e})] c^{2}$

= $[2 \times 12.0 - 19.992439 - 4.002603] c^{2}$

=4.958 $\times 10^{- 3} c^{2}$ u

=4.958 $\times 10^{- 3} \times 931.5$

=4.6183 MeV

The positive sign shows that the reaction is exothermic.

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