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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 eVB: –3.4 eV, –6.8 eVC: 3.4 eV, –6.8 eVD: 3.4 eV, 3.4 eVAns: **C

Solution:

In Bohr's model of H atom

∴ K.E. = 3.4 eV

U = –6.8 eV

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:

Solution:

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

A: 1 : 1

B: 1 : –1

C: 2 : –1

D: 1 : –2

Ans:

Solution:

**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} CB: 10^{–37} CC: 10^{–47} CD: 10^{–20} CAns: BSolution:**

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 ,

gravitational force,

where M

a/c to question,

gravitational force = electrostatic force

we know, K = 9 × 10

so, 9 × 10

or, ∆e

or, ∆e = 1.43767 × 10

A: 1

B: 4

C: 0.5

D: 2

Ans:

Solution:

For last line of Balmer : n

For last line Lyman series : n

A: m

B: 1/m

C: 1/√m

D: 1/m

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,

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,

This is the required distance of closest approach to alpha particle from the nucleus.

A: 2.5 x 10

B: 0.025 x 10

C: 0.5 x 10

D: 0.25 x 10

Ans:

Rydberg constant, r = 10

For last line in Balmer series,

n

We know,

A: 3.0 x 10

B: 2.92 x 10

C: 1.46 x 10

D: 0.73 x 10

Ans: C

Solution:

**Q.9. ****If radius of thenucleus is taken to be R _{Al}, then the radius ofnucleus is nearly : (2015)A:**

B:

C:

D:

**Ans: **C

Solution:

Radius of the nucleus goes as R ∝ A

**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: 6B: 10C: 3D: 2Ans: ASolution:**

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