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## Tuesday, December 30, 2008

### AP Physics B – Nuclear Physics- Multiple Choice Practice Questions

Let us discuss some multiple choice questions on nuclear physics which will benefit AP Physics B aspirants:

(1) The ratio of the nuclear radius of 52Te125 to that of 13Al27 is

(a) 4

(b) 125/17

(c) ¼

(d) 5/3

(e) 3/5

The nuclear radius of an atom of mass number A is given by

R =R0A1/3 where R0 = 1.2×10–15 m.

The required ratio is therefore (125/27)1/3 which is equal to 5/3.

(2) A nucleus of 92U238 gets converted into a 91Pa234 nucleus. The particles emitted during this decay are

(a) one α-particle and one positron

(b) one α-particle and one electron

(c) one α-particle and one antineutrino

(d) one α-particle and one neutrino

(e) one α-particle and two β-particles

The mass number decreases by 4 and the atomic number decreases by 1 in the above decay. When an α-particle is emitted, the mass number decreases by 4 and the atomic number decreases by 2. The atomic number can be increased by one from this condition only if an electron also is emitted. The correct option therefore is (b).

(3) The de Broglie wave length of an α-particle of mass m emitted by a nucleus of mass M initially at rest is λ. The de Broglie wave length of the nucleus immediately after the α-emission is

(a) λ

(b) λ (M m)/m

(c) λm/(Mm)

(d) λ (m/M)2

(e) λ (M/m)2

The nucleus has a recoil momentum on emitting the α-particle. Since the parent nucleus is initially at rest, its recoil momentum has the same magnitude as that of the α-particle but the direction is opposite in accordance with the law of conservation of momentum. The de Broglie wave length λ of the α-particle is given by

λ = h/p where h is Planck’s constant.

Since the recoil momentum of the nucleus has the same magnitude p the de Broglie wave length of the nucleus immediately after the α-emission is λ itself [Option (a)].

(4) Complete the following relation representing one possible fission process:

0n1 + 92U23538Sr90 + ----

(a) 54Xe145

(b) 54Xe145 + 3 0n1

(c) 54Xe143 + 3 0n1

(d) 54Xe142 + 0n1

(e) 54Xe142 + 3 0n1

The total mass number and the total atomic number on the two sides will match only if the relation is completed with the terms given in option (c).

(5) The mass m of any particle of rest mass m0 at speed v is given by Einstein’s relativistic relation,

m = m0/[1– (v2/c2)]1/2 where c is the speed of light in free space.

The rest energy of an electron is 0.511 MeV. The increase in the energy of the electron when it is accelerated from rest to 80% of the speed of light in free space is (very nearly)

(a) 0.341 MeV

(b) 0.405 MeV

(c) 0.511 Mev

(d) 0.852 Mev

(e) 0.916 Mev

We have m0c2 = 0.511 Mev for the electron.

The total energy at a speed of 0.8 c is m0c2/[1– (v2/c2)]1/2 = m0c2/(0.36)1/2 since v = 0.8 c.

But m0c2 = 0.511 Mev so that the total energy is 0.511/0.6 = 0.852 MeV nearly.

The increase in the energy is 0.852 – 0.511 = 0.341 MeV.