16. The mass of a ₃Li⁷ nucleus is 0.042 u less than the sum of the masses of all its nucleons. The binding energy per nucleon of ₃Li⁷ nucleus is nearly​​

1. 1. ​46 MeV ​

   
   
   

   2. 5.6 MeV ​

   
   
   

   3. 3.9 MeV ​

   
   
   

   4. 23 MeV

   
   
   

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1. View Hint View Answer Discuss in Forum

   B.E. = 0.042 × 931 ≃ 42 MeV ​
   Number of nucleons in
   

   	7	Li is 7.
   	3

   
   ∴ ​B.E./ nucleon = 42/7 = 6 MeV ≃ 5.6 MeV

   Correct Option: B

   B.E. = 0.042 × 931 ≃ 42 MeV ​
   Number of nucleons in
   

   	7	Li is 7.
   	3

   
   ∴ ​B.E./ nucleon = 42/7 = 6 MeV ≃ 5.6 MeV

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17. Fusion reaction takes place at high temperature because

1. 1. ​nuclei break up at high temperature

   
   
   

   2. atoms get ionised at high temperature

   
   
   

   3. ​kinetic energy is high enough to overcome the coulomb repulsion between nuclei

   
   
   

   4. molecules break up at high temperature

   
   
   

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1. View Hint View Answer Discuss in Forum

   When the coulomb repulsion between the nuclei is overcome then nuclear fusion reaction takes place. This is possible when temperature is too high.

   Correct Option: C

   When the coulomb repulsion between the nuclei is overcome then nuclear fusion reaction takes place. This is possible when temperature is too high.

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18. A nucleus of mass M emits a photon of frequency ν and the nucleus recoils. The recoil energy will be

1. 1. Mc² – hν ​

   
   
   

   2. h²ν² / 2Mc²

   
   
   

   3. zero

   
   
   

   4. hν

   
   
   

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1. View Hint View Answer Discuss in Forum

   Momentum
   

   Mu =	E	=	hv
   	c		c

   
   Recoil energy
   

   Correct Option: B

   Momentum
   

   Mu =	E	=	hv
   	c		c

   
   Recoil energy
   

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19. The power obtained in a reactor using U²³⁵ disintegration is 1000 kW. The mass decay of U²³⁵ per hour is​​​

1. 1. 10 micro gram ​

   
   
   

   2. 20 micro gram ​

   
   
   

   3. 40 micro gram

   
   
   

   4. 1 micro gram

   
   
   

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1. View Hint View Answer Discuss in Forum

   E = mc²
   

   m =	E
   	c2

   
   So, mass decay per second
   

   	dm	=	1		dE	=	1	(Power in watt)
   	dt		c2		dt		c2

   
   

   =	1	× 1000 × 103
   	(3 × 108)2

   
   and mass decay per hour
   

   =	dm	× 60 × 60
   	dt

   
   

   =	1	× 106 × 3600
   	(3 × 108)2

   
   = 4 × 10^–8 kg ​= 40 micro gram

   Correct Option: C

   E = mc²
   

   m =	E
   	c2

   
   So, mass decay per second
   

   	dm	=	1		dE	=	1	(Power in watt)
   	dt		c2		dt		c2

   
   

   =	1	× 1000 × 103
   	(3 × 108)2

   
   and mass decay per hour
   

   =	dm	× 60 × 60
   	dt

   
   

   =	1	× 106 × 3600
   	(3 × 108)2

   
   = 4 × 10^–8 kg ​= 40 micro gram

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20. How does the binding energy per nucleon vary with the increase in the number of nucleons?

1. 1. ​Increases continuously with mass number

   
   
   

   2. Decreases continuously with mass number ​

   
   
   

   3. First decreases and then increases with increase in mass number

   
   
   

   4. First increases and then decreases with increase in mass number

   
   
   

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1. View Hint View Answer Discuss in Forum

   
   From the graph of BE/A versus mass number A  it is clear that, BE/A first increases and then decreases with increase in mass number.

   Correct Option: D

   
   From the graph of BE/A versus mass number A  it is clear that, BE/A first increases and then decreases with increase in mass number.

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