SATHEE: Atoms And Nuclei Question 159

Atoms And Nuclei Question 159

Question: The binding energy of deuteron $_{1}^{2} H$ is 1.112 MeV per nucleon and an $α -$ particle $_{2}^{4} H e$ has a binding energy of 7.047 MeV per nucleon. Then in the fusion reaction $_{1}^{2} H +_{1}^{2} H \to_{2}^{4} H e + Q$ , the energy Q released is [MP PMT 1992; Roorkee 1994; IIT 1996; AIIMS 1997]

Options:

A) 1 MeV

B) 11.9 MeV

C) 23.8 MeV

D) 931 MeV

Show Answer

Answer:

Correct Answer: C

Solution:

Mass of $_{1} H^{2} = 2.01478 a . m . u .$

Mass of $_{2} H e^{4} = 4.00388 a . m . u .$

Mass of two deuterium $= 2 \times 2.01478 = 4.02956$

Energy equivalent to $2_{1} H^{2}$

$= 4.02956 \times 1.112 M e V = 4.48 M e V$

Energy equivalent to $_{2} H^{4}$

$= 4.00388 \times 7.047 M e V = 28.21 M e V$

Energy released $= 28.21 - 4.48 = 23.73 M e V$ = 24 MeV

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Atoms And Nuclei Question 159

Question: The binding energy of deuteron 12H is 1.112 MeV per nucleon and an α− particle 24He has a binding energy of 7.047 MeV per nucleon. Then in the fusion reaction 12H+12H→24He+Q , the energy Q released is [MP PMT 1992; Roorkee 1994; IIT 1996; AIIMS 1997]

Options:

Answer:

Solution:

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