21. An alpha nucleus of energy
    

    	1	mv2
    	2

    
    bombards a heavy nuclear target of charge Ze. Then the distance of closest approach for the alpha nucleus will be proportional to

1. 1. 	1
      	Ze

   
   
   

   2. v²

   
   
   

   3. ​

      	1
      	m

   
   
   

   4. 	1
      	v4

   
   
   

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1. View Hint View Answer Discuss in Forum

   Kinetic energy of alpha nucleus is equall to electrostatic potential energy of the system of the alpha particle and the heavy nucleus. That is,
   

   	1	mv2 =	1		qα Ze
   	2		4πε0		r0

   
   where r₀ is the distance of closest approach
   

   r0 =	2		qα Ze
   	4πε0		mv2

   
   

   ⇒ r0 ∝ Ze ∝ qα ∝	1	∝	1	= 8
   	m		v2

   
   Hence, correct option is (c).

   Correct Option: C

   Kinetic energy of alpha nucleus is equall to electrostatic potential energy of the system of the alpha particle and the heavy nucleus. That is,
   

   	1	mv2 =	1		qα Ze
   	2		4πε0		r0

   
   where r₀ is the distance of closest approach
   

   r0 =	2		qα Ze
   	4πε0		mv2

   
   

   ⇒ r0 ∝ Ze ∝ qα ∝	1	∝	1	= 8
   	m		v2

   
   Hence, correct option is (c).

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22. When an α-particle of mass 'm' moving with velocity 'v' bombards on a heavy nucleus of charge 'Ze', its distance of closest approach from the nucleus depends on m as :​

1. 1. 	1
      	m

   
   
   

   2. ​

      	1
      	√m

   
   
   

   3. 	1
      	m2

   
   
   

   4. m

   
   
   

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1. View Hint View Answer Discuss in Forum

   At closest distance of approach, the kinetic energy of the particle will convert completely into electrostatic potential energy. ​
   Kinetic energy
   

   K.E. =	1	mv2
   	2

   
   Potential energy
   

   P.E. =	KQq
   	r

   
   

   	1	mv2 =	KQq	⇒ r ∝	1
   	2		r		m

   Correct Option: A

   At closest distance of approach, the kinetic energy of the particle will convert completely into electrostatic potential energy. ​
   Kinetic energy
   

   K.E. =	1	mv2
   	2

   
   Potential energy
   

   P.E. =	KQq
   	r

   
   

   	1	mv2 =	KQq	⇒ r ∝	1
   	2		r		m

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23. Which of the following transitions in a hydrogen atom emits the photon of highest frequency?

1. 1. n = 2 to n = 1 ​

   
   
   

   2. n = 2 to n = 6

   
   
   

   3. n = 6 to n = 2 ​

   
   
   

   4. n = 1 to n = 2

   
   
   

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1. View Hint View Answer Discuss in Forum

   V = R		1	-	1
   		n21		n22

   
   or,

   1	= R		1	-	1
   λ			n21		n22

   
   Frequency,
   

   v = Rc		1	-	1
   		n21		n22

   
   Note : See the greatest energy difference and also see that the transition is from higher to lower energy level. Hence, it is highest in case of n = 2 to n = 1.

   Correct Option: A

   V = R		1	-	1
   		n21		n22

   
   or,

   1	= R		1	-	1
   λ			n21		n22

   
   Frequency,
   

   v = Rc		1	-	1
   		n21		n22

   
   Note : See the greatest energy difference and also see that the transition is from higher to lower energy level. Hence, it is highest in case of n = 2 to n = 1.

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24. To explain his theory, Bohr used

1. 1. conservation of linear momentum

   
   
   

   2. conservation of angular momentum

   
   
   

   3. conservation of quantum frequency

   
   
   

   4. conservation of energy

   
   
   

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1. View Hint View Answer Discuss in Forum

   Bohr used conservation of angular momentum.

   Correct Option: B

   Bohr used conservation of angular momentum.

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25. The ground state energy of H-atom 13.6 eV. The energy needed to ionize H-atom from its second excited state.

1. 1. 1.51 eV

   
   
   

   2. 3.4 eV

   
   
   

   3. 13.6 eV

   
   
   

   4. 12.1 eV

   
   
   

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1. View Hint View Answer Discuss in Forum

   Second excited state corresponds to n = 3
   

   ∴ E =	13.6	eV = 1.51 eV
   	32

   Correct Option: A

   Second excited state corresponds to n = 3
   

   ∴ E =	13.6	eV = 1.51 eV
   	32

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