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Standard XII

Physics

Nuclear Fusion

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 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

Q. Calculate the binding energy per nucleon for

_{10}^{20} N e,_{26}^{56} F e

and

_{92}^{238} U

. Given that mass of neutron is 1.008665 amu, mass of proton is 1.007825 amu, mass of

_{10}^{20} N e

is 19.9924 amu, mass of

_{26}^{56} F e

is 55.93492 amu,

_{92}^{238} U

is 238.050783 amu

Q. A heavy nucleus

X

of mass number

240

and binding energy per nucleon

7.6 M e

is split into two fragment

Y

and

Z

of mass numbers

110

and

130

. The binding energy per nucleon in

Y

and

Z

8.5 M e V

per nucleon. Calculate the energy

Q

released per fission in

M e V

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