71. The count rate of a Geiger Muller counter for the radiation of a radioactive material of half-life 30 minutes decreases to 5 sec^–1 after 2 hours. The initial count rate was

1. 1. 20 sec^–1

   
   
   

   2. 25 sec^–1

   
   
   

   3. 80 sec^–1

   
   
   

   4. 625 sec^–1

   
   
   

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1. View Hint View Answer Discuss in Forum

   Half-life = 30 minutes; Rate of decrease (N) = 5 per second and total time = 2 hours = 120 minutes. Relation for initial and final count rate
   

   	N	=		1		time/half-life
   	N0			2

   
   

   =		1		120/30
   		2

   
   

   =		1		4	=	1
   		2				16

   
   Therefore, N₀ = 16 × N = 16 × 5 = 80 s^–1.

   Correct Option: C

   Half-life = 30 minutes; Rate of decrease (N) = 5 per second and total time = 2 hours = 120 minutes. Relation for initial and final count rate
   

   	N	=		1		time/half-life
   	N0			2

   
   

   =		1		120/30
   		2

   
   

   =		1		4	=	1
   		2				16

   
   Therefore, N₀ = 16 × N = 16 × 5 = 80 s^–1.

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72. The half life of radium is 1600 years. The fraction of a sample of radium that would
    remain after 6400 years

1. 1. 1/4

   
   
   

   2. 1/2

   
   
   

   3. 1/8

   
   
   

   4. 1/16

   
   
   

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1. View Hint View Answer Discuss in Forum

   	N	=		1		6400/1600
   	N0			2

   
   

   =		1		4	=	1
   		2				16

   Correct Option: D

   	N	=		1		6400/1600
   	N0			2

   
   

   =		1		4	=	1
   		2				16

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73. The nucleus ₆C¹² absorbs an energetic neutron and emits a beta particle (β). The resulting nucleus is

1. 1. ₇N¹⁴

   
   
   

   2. ₇N¹³

   
   
   

   3. ₅B¹³

   
   
   

   4. ₆C¹³

   
   
   

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1. View Hint View Answer Discuss in Forum

   ₆C¹² + ₀n¹ → ₆C¹³ → ₇N¹³ + _-1β⁰ + Energy

   Correct Option: B

   ₆C¹² + ₀n¹ → ₆C¹³ → ₇N¹³ + _-1β⁰ + Energy

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74. A mixture consists of two radioactive materials A₁ and A₂ with half lives of 20 s and 10 s respectively. Initially the mixture has 40 g of A₁ and 160 g of A₂ . The amount of the two in the mixture will become equal after :

1. 1. 60 s

   
   
   

   2. 80 s

   
   
   

   3. 20 s

   
   
   

   4. 40 s

   
   
   

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1. View Hint View Answer Discuss in Forum

   Let, the amount of the two in the mixture will become equal after t years.
   The amount of A₁, which remains after t years
   

   N1 =	N01
   	(2)t/20

   
   The amount of A₂, which remains, after t years
   

   N2 =	N02
   	(2)t/10

   
   According to the problem
   N₁ = N₂
   

   	40	=	160
   	(2)t/20		(2)t/10

   
   2^t/20 = 2^{(t/10) - 2}
   

   	t	=	t	- 2
   	20		10

   
   

   	t	-	t	= 2
   	20		10

   
   

   	t	= 2
   	200

   
   t = 40 s

   Correct Option: D

   Let, the amount of the two in the mixture will become equal after t years.
   The amount of A₁, which remains after t years
   

   N1 =	N01
   	(2)t/20

   
   The amount of A₂, which remains, after t years
   

   N2 =	N02
   	(2)t/10

   
   According to the problem
   N₁ = N₂
   

   	40	=	160
   	(2)t/20		(2)t/10

   
   2^t/20 = 2^{(t/10) - 2}
   

   	t	=	t	- 2
   	20		10

   
   

   	t	-	t	= 2
   	20		10

   
   

   	t	= 2
   	200

   
   t = 40 s

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75. Energy released in the fission of a single ₉₂U²³⁵ nucleus is 200 MeV. The fission rate of a ₉₂U²³⁵ filled reactor operating at a power level of 5 W is

1. 1. 1.56 × 10^–10 s^–1

   
   
   

   2. 1.56 × 10¹¹ s^–1

   
   
   

   3. 1.56 × 10^–16 s^–1

   
   
   

   4. 1.56 × 10^–17 s^–1

   
   
   

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1. View Hint View Answer Discuss in Forum

   Fission rate
   

   =	total power	=	5
   	energy		200 × 1.6 × 10-13
   	fission

   
   = 1.56 × 10¹¹ s^–1

   Correct Option: B

   Fission rate
   

   =	total power	=	5
   	energy		200 × 1.6 × 10-13
   	fission

   
   = 1.56 × 10¹¹ s^–1

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