1 Marks Question

Question 11 Mark

³²P beta-decays to ³²S. Find the sum of the energy of the antineutrino and the kinetic energy of the $\beta$-particle. Neglect the recoil of the daughter nucleus. Atomic mass of ³²P = 31.974u and that of ³²S = 31.972u.

Answer

$\text{P}^{32}\rightarrow\text{S}^{32}+ \ _0\bar{\text{v}}^0+ \ _1\beta^0$
Energy of antineutrino and $\beta$-particle
= (31.974 - 31.972)u = 0.002u = 0.002 × 931 = 1.862MeV = 1.86.

View full question & answer→

Question 21 Mark

Calculate the minimum energy needed to separate a neutron from a nucleus with Z protons and N neutrons it terms of the masses M_Z.N, M_Z_,N-1 and the mass of the neutron.

Answer

$\text{E}_2\text{N}=\text{E}_{\text{Z,N}-1}+\text{ }^1_0\text{n}.$
Energy released = (Initial Mass of nucleus - Final mass of nucleus)c² = (M_Z.N-1 + M₀ - M_ZN)c².

View full question & answer→

Question 31 Mark

Show that the minimum energy needed to separate a proton from a nucleus with Z protons and N neutrons is:
$\Delta\text{E}=(\text{M}_{\text{Z}-1,\text{N}}+\text{M}_{\text{H}}-\text{M}_{\text{Z,N}})\text{c}^2$
where M_Z,N = mass of an atom with Z protons and N neutrons in the nucleus and M_H = mass of a hydrogen atom. This energy is known as proton-separation energy.

Answer

E_Z.N. → E_Z-1, N + P₁ ⇒ E_Z.N. → E_Z-1, N + ₁H¹ [As hydrogen has no neutrons but protons only]
$\Delta$E = (M_Z-1,N + N_H - M_Z,N)c²

View full question & answer→

Question 41 Mark

Calculate the energy released if ²³⁸U emits an $\alpha$-particle.
Calculate the energy to be supplied to ²³⁸U it two protons and two neutrons are to be emitted one by one. The atomic masses of ²³⁸U, ²³⁴Th and ⁴He are 238.0508u, 234.04363u and 4.00260u respectively.

Answer

U²³⁸ ₂He⁴ + Th²³⁴

E = [Mu - (N_HC + M_Th)]u = 238.0508 - (234.04363 + 4.00260)]u = 4.25487Mev = 4.255Mev.

E = U²³⁸ - [Th²³⁴ + 2n'₀ + 2p'₁]

= {238.0508 - [234.64363 + 2(1.008665) + 2(1.007276)]}u

= 0.024712u = 23.0068 = 23.007MeV.

View full question & answer→

Physics 1 Marks Question

Test yourself on this topic