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Physics 28 Years Past year papers for NEET/AIPMT Class 12

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The document Notes | EduRev is a part of the NEET Course Physics 28 Years Past year papers for NEET/AIPMT Class 12.
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Q.1. The total energy of an electron in an atom in an orbit is –3.4 eV. Its kinetic and potential energies are, respectively:    (2019)
A: –3.4 eV, –3.4 eV
B: –3.4 eV, –6.8 eV
C: 3.4 eV, –6.8 eV
D: 3.4 eV, 3.4 eV
Ans: 
C
Solution:

In Bohr's model of H atom
 Notes | EduRev
∴ K.E. = 3.4 eV
U = –6.8 eV

Q.2. α-particle consists of :    (2019)
A: 2 protons and 2 neutrons only
B: 2 electrons, 2 protons and 2 neutrons
C: 2 electrons and 4 protons only
D: 2 protons only
Ans:
A
Solution:

α-particle is nucleus of Helium which has two protons and two neutrons.

Q.3. The ratio of kinetic energy to the total energy of an electron in a Bohr orbit of the hydrogen atom,is :-    (2018)
A: 1 : 1
B: 1 : –1
C: 2 : –1
D: 1 : –2
Ans: 
B
Solution:

 Notes | EduRev

Q.4. Suppose the charge of a proton and an electron differ slightly. One of them is – e, the other is(e + De). If the net of electrostatic force and gravitational force between two hydrogen atoms placed at a distance d (much greater than atomic size) apart is zero, then De is of the order of [Given mass of hydrogen mh = 1.67 × 10–27 kg]     (2017)
A: 10–23 C
B: 10–37 C
C: 10–47 C
D: 10–20 C
Ans: B
Solution:

we know, a hydrogen atom has one electron and one proton
so, net charge on each hydrogen atom = (e + ∆e - e) = ∆e
so, electrostatic force ,  Notes | EduRev
gravitational force,  Notes | EduRev
where Mh denotes mass of hydrogen
a/c to question,
gravitational force = electrostatic force
 Notes | EduRev
 Notes | EduRev
we know, K = 9 × 109 Nm²/C², G = 6.67 × 10-11 Nm²/Kg² , Mh = 1.67 × 10-27 Kg
so, 9 × 109 × ∆e² = 6.67 × 10-11 × (1.67 × 10-27)²
or, ∆e² = 6.67 × 10-11 × 1.67 × 1.67 × 10-54/9 × 109
or, ∆e = 1.43767 × 10-37 C

Q.5. The ratio of wavelengths of the last line of Balmer series and the last line of Lyman series is :-     (2017)
A: 1
B: 4
C: 0.5
D: 2
Ans: 
B
Solution:

For last line of Balmer : n1 = 2 & n2 = ∞
 Notes | EduRev
For last line Lyman series : n1 = 1 & n2 = ∞
 Notes | EduRev
 Notes | EduRev

Q.6. When an α-particle of mass 'm' moving with velocity 'v' bombards on a heavy nucleus of charge 'Ze' its distance of closet approach from the nucleus depends on m as :    (2016)
A: m
B: 1/m
C: 1/√m
D: 1/m2
Ans: B
Solution:

When an alpha particle moving with velocity v bombards on a heavy nucleus of charge Ze, then there will be no loss of energy.
Initial Kinetic energy of the alpha particle = Potential energy of alpha particle at closest approach.
That is,
 Notes | EduRev
 Notes | EduRev
This is the required distance of closest approach to alpha particle from the nucleus.
When an alpha particle moving with velocity v bombards on a heavy nucleus of charge Ze, then there will be no loss of energy.
Initial Kinetic energy of the alpha particle = Potential energy of alpha particle at closest approach.
That is,
 Notes | EduRev
 Notes | EduRev
This is the required distance of closest approach to alpha particle from the nucleus.

Q.7. Given the value of Rydberg constant is 107 m-1, the wave number of the last line of the Balmer series in hydrogen spectrum will be :     (2016)
A: 2.5 x 107m-1
B: 0.025 x 104m-1
C: 0.5 x 107m-1
D: 0.25 x 107m-1
Ans: 
D
Solution:
Rydberg constant, r = 107 m-1
For last line in Balmer series,
n2 = ∞; n1 = 2
We know,
 Notes | EduRev

Q.8. Consider 3rd orbit of He+ (Helium), using non-relativistic approach, the speed of electron in this orbit will be [given K = 9 x 109 constant, Z = 2 and h(Planck's Constant) = 6.6 x 10-34 J s]    (2015)
A: 3.0 x 108 m/s
B: 2.92 x 106 m/s
C: 1.46 x 106 m/s
D: 0.73 x 106 m/s
Ans: C
Solution:

 Notes | EduRev

Q.9. If radius of the Notes | EduRevnucleus is taken to be RAl, then the radius of Notes | EduRevnucleus is nearly :    (2015)
A:

 Notes | EduRev
B:

 Notes | EduRev
C:

 Notes | EduRev
D:

 Notes | EduRev
Ans: 
C
Solution:

Radius of the nucleus goes as R ∝ A1/3, where A is the atomic mass. If RTe is the radius of the nucleus of telurium atom and RAl is the radius of the nucleus of aluminium atom we have

 Notes | EduRev

Q.10. Hydrogen atom in ground state is excited by a monochromatic radiation of λ = 975 Å. Number of spectral lines in the resulting spectrum emitted will be:     (2014)
A: 6
B: 10
C: 3
D: 2
Ans: A
Solution:

 Notes | EduRev
 Notes | EduRev

 Notes | EduRev

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