Fusion reaction is given by 126C - 126C - 2010Ne - 42He-If B-E- per nucleon for 126C is X- B-E- per nucleon for 2010Ne is Y and B-E- per nucleon for 42He is Z- then Q value of reaction is

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Standard XII

Chemistry

Mass Energy Equivalence

Fusion reacti...

Question

Fusion reaction is given by $_{6}^{12} C +_{6}^{12} C \to_{10}^{20} N e +_{2}^{4} H e$ .
If B.E. per nucleon for $_{6}^{12} C$ is X, B.E. per nucleon for $_{10}^{20} N e$ is Y and B.E. per nucleon for $_{2}^{4} H e$ is Z, then Q value of reaction is

24 X - 20 Y - 6 Z

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Y + Z - X

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24 X - 10 Y - 2 Z

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20 Y + 4 Z - 24 X

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Solution

The correct option is D $20 Y + 4 Z - 24 X$
$Q = B . E ._{p r o d} - B . E ._{r e a c t a n t}$
$Q = 20 Y + 4 Z - 24 X$

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Similar questions

Q. Fusion reaction is given by

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

.
If B.E. per nucleon for

_{6}^{12} C

is X, B.E. per nucleon for

_{10}^{20} N e

is Y and B.E. per nucleon for

_{2}^{4} H e

is Z, then Q value of reaction is

Q. The Q value of a nuclear reaction, A + b

\to

C + d, is defined by

Q = [m_{A} + m_{b}] - [m_{C} + m_{d}]

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 +_{3}^{1} 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

Q. Calculate mass defect and binding energy per nucleon of

_{10}^{20} N e,

given
Mass of

_{10}^{20} N e = 19.992397

u
Mass of

_{1}^{1} H = 1.007825

u
Mass of

_{0}^{1} n = 1.008665

Q. The Binding energy per nucleon of

_{3}^{7} L i

and

_{2}^{4} H e

nucleon are

5.60 M e V

and

7.06 M e V

, respectively. In the nuclear reaction

_{3}^{7} L i +_{1}^{1} H \to_{2}^{4} H e +_{2}^{4} H e + Q

, the value of energy

Q

released is

The binding energy of deuteron $(_{1}^{2} H)$ is 1.15 MeV per nucleon and an alpha particle $(_{2}^{4} H e)$ has a binding energy of 7.1 MeV per nucleon. Then in the reaction
$_{1}^{2} H +_{1}^{2} H \to_{2}^{4} H e + Q$
the energy Q released is

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Fusion reaction is given by 126C+126C→2010Ne+42He. If B.E. per nucleon for 126C is X, B.E. per nucleon for 2010Ne is Y and B.E. per nucleon for 42He is Z, then Q value of reaction is

The correct option is D 20Y+4Z−24XQ=B.E.prod−B.E.reactant Q=20Y+4Z−24X

Fusion reaction is given by $_{6}^{12} C +_{6}^{12} C \to_{10}^{20} N e +_{2}^{4} H e$ .
If B.E. per nucleon for $_{6}^{12} C$ is X, B.E. per nucleon for $_{10}^{20} N e$ is Y and B.E. per nucleon for $_{2}^{4} H e$ is Z, then Q value of reaction is

The correct option is D $20 Y + 4 Z - 24 X$
$Q = B . E ._{p r o d} - B . E ._{r e a c t a n t}$
$Q = 20 Y + 4 Z - 24 X$