26. The ionization energy of hydrogen atom is 13.6 eV. Following Bohr’s theory, the energy corresponding to a transition between 3rd and 4th orbit is

1. 1. 3.40 eV

   
   
   

   2. 1.51 eV

   
   
   

   3. 0.85 eV

   
   
   

   4. 0.66 eV

   
   
   

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1. View Hint View Answer Discuss in Forum

   E = E₄ – E₃
   

   = -	13.6	-		-	13.6		= - 0.85 + 1.51 = 0.66 eV
   	42				32

   Correct Option: D

   E = E₄ – E₃
   

   = -	13.6	-		-	13.6		= - 0.85 + 1.51 = 0.66 eV
   	42				32

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27. In terms of Bohr radius a₀, the radius of the second Bohr orbit of a hydrogen atom is given by

1. 1. 4 a₀

   
   
   

   2. 8 a₀

   
   
   

   3. √2a₀

   
   
   

   4. 2 a₀

   
   
   

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1. View Hint View Answer Discuss in Forum

   As r ∝ n² therefore, radius of 2nd Bohr’s orbit = 4 a₀.

   Correct Option: A

   As r ∝ n² therefore, radius of 2nd Bohr’s orbit = 4 a₀.

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28. Which source is associated with a line emission spectrum ?

1. 1. Electric fire

   
   
   

   2. Neon street sign

   
   
   

   3. Red traffic light

   
   
   

   4. Sun

   
   
   

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1. View Hint View Answer Discuss in Forum

   Neon street sign is a source of line emission spectrum.

   Correct Option: B

   Neon street sign is a source of line emission spectrum.

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29. The radius of hydrogen atom in its ground state is 5.3 × 10^–11 m. After collision with an electron it is found to have a radius of 21.2 × 10^–11 m. What is the principal quantum number n of the final state of the atom

1. 1. n = 4

   
   
   

   2. n = 2

   
   
   

   3. n = 16

   
   
   

   4. n = 3

   
   
   

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1. View Hint View Answer Discuss in Forum

   r ∝ n²
   

   ∴	radius of final state	= n2
   	radius of initial state

   
   

   	21.2 × 10-11	= n2
   	5.3 × 10-11

   
   ∴ n² = 4 or n = 2

   Correct Option: B

   r ∝ n²
   

   ∴	radius of final state	= n2
   	radius of initial state

   
   

   	21.2 × 10-11	= n2
   	5.3 × 10-11

   
   ∴ n² = 4 or n = 2

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30. When a hydrogen atom is raised from the ground state to an excited state,

1. 1. P.E decreases and K.E. increases

   
   
   

   2. P.E. increases and K.E decreases

   
   
   

   3. both K.E. and P.E. decrease

   
   
   

   4. absorption spectrum

   
   
   

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1. View Hint View Answer Discuss in Forum

   P.E = -	KZe2	and K.E =	KZe2
   	r		2r

   
   where, r is the radius of orbit which increases as we move from ground to an excited state. Therefore, when a hydrogen atom is raised from the ground state, it increases the value of r. As a result of this, P.E. increases (decreases in negative) and K.E. decreases.

   Correct Option: B

   P.E = -	KZe2	and K.E =	KZe2
   	r		2r

   
   where, r is the radius of orbit which increases as we move from ground to an excited state. Therefore, when a hydrogen atom is raised from the ground state, it increases the value of r. As a result of this, P.E. increases (decreases in negative) and K.E. decreases.

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