1 Marks Question

Question 11 Mark

${ }^{32} \mathrm{P}$ beta-decays to ${ }^{32} \mathrm{~S}$. 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 ${ }^{32} \mathrm{P}=31.974 \mathrm{u}$ and that of ${ }^{32} \mathrm{~S}=31.972 \mathrm{u}$.

Answer

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

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Question 21 Mark

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

Answer

a. $\mathrm{U}^{238}{ }_2 \mathrm{He}^4+\mathrm{Th}^{234}$
$E= {\left.\left[M u-\left(N_{H C}+M_{T h}\right)\right] u=238.0508-(234.04363+4.00260)\right] u=4.25487 \mathrm{Mev}=4.255 \mathrm{Mev} . }$
$\quad \text { b. } E=U^{238}-\left[T h^{234}+2 n_0^{\prime}+2 p_1^{\prime}\right]$
=$\{238.0508-[234.64363+2(1.008665)+2(1.007276)]\} u$
=$0.024712 u=23.0068=23.007 \mathrm{MeV} .$

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Question 31 Mark

Calculate the minimum energy needed to separate a neutron from a nucleus with Z protons and N neutrons it terms of the masses $\mathrm{M}_{\mathrm{Z}, \mathrm{N}}, \mathrm{M}_{\mathrm{Z}, \mathrm{N}-1}$ 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 $) \mathrm{c}^2=\left(\mathrm{M}_{\mathrm{Z}. \mathrm{N}-1}+\mathrm{M}_0-\mathrm{M}_{\mathrm{ZN}}\right) \mathrm{c}^2$.

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Question 41 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

$\mathrm{E}_{\mathrm{ZN} .} \rightarrow \mathrm{E}_{\mathrm{Z}-1}, \mathrm{~N}+\mathrm{P}_1 \Rightarrow \mathrm{E}_{\mathrm{ZN}} \rightarrow \mathrm{E}_{\mathrm{Z}-1}, \mathrm{~N}+{ }_1 \mathrm{H}^1\left[\right.$ As hydrogen has no neutrons but protons only] $\Delta \mathrm{E}=\left(\mathrm{M}_{\mathrm{Z}-1}, \mathrm{~N}+\mathrm{N}_{\mathrm{H}}-\right.$ $\left.\mathrm{M}_{\mathrm{Z}, \mathrm{N}}\right) \mathrm{c}^2$

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