Physics 3 Marks Question

1. The reaction invoved is,

$^{226}_{88}\text{Ra}\ \rightarrow^{222}_{86}\text{Rn }\ + \ ^{4}_{2}\text{He}$

The difference in mass between the original nucleus and the decay products = 226.02540 u - (222.01750 u + 4.00260 u)

= + 0.0053 u

$\therefore\ $Energy equivalent or Q-value = 0.0053 × 931.5 MeV

= 4.93695 MeV

= 4.94 MeV

The decay products would emerge with total kinetic energy 4.94 MeV.

Momentum is conserved. If the parent nucleus is at rest, the daughter and the a-particle have momenta of equal magnitude p but, in the opposite direction.

Kinetic energy, $\text{K}=\frac{\text{p}^2}{2\text{m}}.$

Since, p is the same for the two particles therefore the kinetic energy divides inversely as their masses.

The $\alpha$-particle gets $\frac{222}{222+4}$ of the total i.e., $\frac{222}{226}\times4.94\text{ MeV }\text{ or }4.85\text{ MeV}.$

2. The difference in mass between the original nucleus and the decay products = 220.01137 u - (216.00189 u + 4.00260 u)

= 0.00688 u

$\therefore\ $Q-value or Energy equivalent = 0.00688 × 931.5 MeV

= 6.41 MeV

Energy of the alpha particle, $\text{E}_\alpha=\frac{216}{216+4}\times6.41\text{ MeV}=6.29\text{ MeV}.$

1. $_{88}\text{Ra}^{226}\xrightarrow{\ \ \ \ \ \ \ } \ _{86}\text{Rn}^{222}+_2\text{He}^4$
   
2. $^{242}_{94}\xrightarrow{\ \ \ \ \ \ \ } \ ^{86}_{92}\text{U }+\ ^4_2\text{He}$
   
3. $^{32}_{15}\text{P}\xrightarrow{\ \ \ \ \ \ \ } \ ^{32}_{16}\text{S }+\ \text{e}^{-}+ _\text{V}^{-}$
   
4. $^{210}_{83}\text{B}\xrightarrow{\ \ \ \ \ \ \ } \ ^{210}_{84}\text{PO }+\ \text{e}^{-}+\ _\text{V}^{-}$
   
5. $^{11}_{6}\text{C}\xrightarrow{\ \ \ \ \ \ \ } \ ^{11}_{5}\text{B }+\ \text{e}^{+}+\ \text{V}$
   
6. $^{97}_{43}\text{Tc}\xrightarrow{\ \ \ \ \ \ \ } \ ^{97}_{42}\text{MO }+\ \text{e}^{+}+ \text{V}$
   
7. $^{120}_{54}\text{Xe}+\text{e}^{+}\xrightarrow{\ \ \ \ \ \ \ } \ ^{120}_{53}\text{I }+\ \text{V}$
   

1. Considering the first reaction,

$^1_1\text{H}\ +\ ^3_1\text{H}\rightarrow\ ^2_1\text{H}\ +\ ^2_1\text{H}$

Q-value is given by,

$\text{Q}=\Delta\text{m}\times931.5\text{ MeV}$

$=[\text{m}(^1_1\text{H})+\text{m}(^3_1\text{H}-2\text{m}(^2_1\text{H})]\times931\text{ MeV}$

$=[1.007825+3.016049-2\times2.014102]\times931\text{ MeV}$

$=-4.03\text{ MeV}$

Since, Q-value is negative, this reaction is endothermic.

2. The second reaction is,

$^{12}_6\text{C}\ +\ ^{12}_6\text{C}\rightarrow\ ^{20}_{10}\text{Ne}\ +\ ^4_2\text{He}$

Q-value is given by,

$\text{Q}=\Delta\text{m}\times931\text{ MeV}$

$=[2\text{m}(^{12}_6\text{C})-\text{m}(^{20}_{10}\text{Ne})-\text{m}(^4_2\text{He})]\times931\text{ MeV}$

$=[24.000000-19.992439-4.002603]\times931\text{ MeV}$

$=+\ 4.61\text{ MeV}$

Since, the Q-value is positive, the reaction is exothermic.

1. Basic nuclear reaction.

$\text{P}\rightarrow\text{n} + \text{e}^{+} + \text{v}$

1. $\text{x} = \beta^{+}/^{o}_{1}\text{e} , \text{y} = 5 , \text{z} =11 $
2. $\text{a} = 10 , \text{b}= 2 , \text{c} = 4.$

Unit – second

1. ${N}(t)=N_0\text{ }e^{-\lambda{t}}$

When $t=T_{1/2}\Rightarrow{N}(t)=\frac{N_0}{2}$

$\therefore {\text{ }} \frac{N_0}{2}=N_0\text{ }e^{-\lambda}T_{1/2}$

$\Rightarrow\frac{1}{2}=e^{-\lambda}T_{1/2}$

$\Rightarrow-{\lambda}T_\frac{1}{2}=-ln2$

$\Rightarrow\text{ }T_\frac{1}{2}=\frac{ln2}{\lambda}$

$\Rightarrow\frac{0.693}{\lambda}$

2. $\frac{N}{N_0}=\bigg(\frac{1}{2}\bigg)^n\text{ }\text{ }\text{ }\text{ }n=\frac{t}{T_{1/2}}$

$\text{Given}\text{ }\frac{N}{N_0}=\frac{1}{4}=\bigg(\frac{1}{2}\bigg)^n $

$\bigg(\frac{1}{2}\bigg)^n=\bigg(\frac{1}{2}\bigg)^2$​​​​​​​

$\therefore \text{Number of half lives}=2$

$\Rightarrow\frac{1000}{T_{1/2}}=2$

$\Rightarrow{T}_\frac{1}{2}=\frac{1000}{2}=500 \text{ }\text{years}$

[𝑨𝒍𝒕𝒆𝒓𝒏𝒂𝒕𝒊𝒗𝒆𝒍𝒚

1000 years = 2 half lives

$\therefore$ Half life = 500 years]

1. $\frac{mv^{2}}{r}=𝑞𝑣𝐵$.

$\therefore{r}=\frac{mv}{𝑞𝐵}=\frac{𝑝}{𝑞𝐵}\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }(p = mv)$

For proton $r_p=\frac{p}{q_p{B}}$

For $\alpha$ particles ${r}_{\alpha}=\frac{p}{q_{\alpha}{B}}$

$\therefore\text{ }\frac{r_p}{r_{\alpha}}=\frac{q_{\alpha}}{q_p}=2$

2. $r=\frac{𝑚𝑣}{𝑞𝐵}=\frac{1}{B}\sqrt{\frac{2𝑚𝑉}{𝑞}}$

for proton $r_p=\frac{1}{B}\sqrt\frac{2m_p{V}}{q_p}$

and for $\alpha$ particles $r_\alpha=\frac{1}{B}\sqrt\frac{2m_{\alpha}{V}}{q_{\alpha}}$

$ \therefore\frac{r_p}{r{\alpha}}=\sqrt{\frac{m_p}{q_p}\frac{q_{\alpha}}{m{\alpha}}}$

$\sqrt{\frac{2}{4}}=\frac{1}{\sqrt{2}}$

1. A nucleus, that spontaneously decays by emitting an electron, or a positron, is said to undergo $\beta$ decay

[Alternatively $ ^{A}_{Z}\text{X}\longrightarrow^{\text{ }\text{ }\text{ }\text{ }\text{ }A}_{Z+1}\text{Y}+e^-+\bar{v}$

$ ^{A}_{Z}\text{X}\longrightarrow^{\text{ }\text{ }\text{ }\text{ }\text{ }A}_{Z-1}\text{Y}+e^++v$ (antineutrino)

During β decay, nucleons undergo transformation. We can have:

${n}\text{ }{\longrightarrow}\text{ }{p}+e^-+{\bar{v}}$

$ \longrightarrow$ A neutron converts into a proton and an electron [Alternatively

${p}\longrightarrow{n}+e^++v$

[A proton converts into a neutron and a positron] It is because the neutrinos, or antineutrino, carry off different amounts of energy.

2. The daughter nuclei have more binding energy per nucleon.

1. Number of radioactive nuclei decaying per second at any time.
2. $R_1=\lambda_1N_1=\frac{0.693}{T_1}N_1$

$R_2=\lambda_2N_2=\frac{0.693}{T_2}N_2$

$\frac{R_1}{R_2}=\frac{N_1}{N_2}\times\frac{T_2}{T_1}$

1. Statement: Rate of decay of a given radioactive sample is directly propotional to the total number of undecayed nuclei present in the sample.

2. $N=N_0e^{-\lambda t}/\frac{N}{N_0}=\Big(\frac{1}{2}\Big)^n$

$\text{n}=\frac{t}{T_{1/2}}=\frac{10}{20}=\frac{1}{2}$

$\Rightarrow N=4\sqrt{2}\times10^{6}\times\Big(\frac{1}{2}\Big)^{\frac{1}{2}}$

$=4\times10^6$ nuclei.

1. $\gamma-rays$

Range: 10¹⁹ to 10²³ Hz

2. To protect the eyes from large amount of UV radiations produced by welding arcs.
3. Because water molecules present in the materials readily absorb the infra red rays get heated up.

$\frac{\text{dN}}{\text{dt}} = -\lambda\text{N}$

$\int_{N_{0}}^{N}\frac{\text{dN}}{\text{N}} = \int_{0}^{t} - \lambda\text{dt}$

$[\log_{e}\text{N}]_{N_{0}}^{N} = - \lambda[\text{t}]_{0}^{t}$

$\log_{e}\frac{\text{N}}{\text{N}_{0}} = - \lambda\text{t}$

$\text{N} = \text{N}_{0}\text{e}^{-\lambda\text{t}}$

1. $^{22}_{11}\text{Na}\rightarrow^{22}_{11}\text{Ne} + \text{e}^{+} + \text{v}$ Also accept, if a student does not identify the product nucleus and writes as

$^{22}_{11}\text{Na}\rightarrow^{22}_{10}\text{X} + \text{e}^{+} + \text{v}$

Basic process

$\text{p}\rightarrow\text{n} + \text{e}^{+} +\text{v}$

2. Isobar.

1. In nuclear reaction

$^{2}_{1}\text{H} +^{2}_{1}\text{H}\rightarrow^{3}_{2}\text{He} + \text{n} + 3.27\text{MeV}$

Cause of the energy released:

1. Binding energy per nucleon of$^{3}_{2}\text{ He}$ becomes more than the (BE/A) of $^{2}_{1}\text{ H}.$
2. Mass defect between the reactant and product nuclei.

$\Delta\text{E} = \Delta\text{mC}^{2}$

$ = \big[2\text{m}(^{2}_{1}\text{H}) - \text{m}(^{3}_{2}\text{He}) + \text{m}(\text{n})\big]\text{C}^{2}$

2. The radius of nucleus of mass number A is given by R = R₀A^1/3

Volume of the nucleus $\text{V} = \frac{4}{3}\pi\text{R}^{3} = \frac{4}{3}\pi\text{R}_{0}^{3}\text{A}$

Density of the matter in the nucleus

$\rho = \frac{\text{Mass}}{\text{Volume}} = \frac{\text{A}{(u)}}{\frac{4}{3}\pi\text{R}_{0}^{3}\text{A}}$

$\rho = \frac{1}{\frac{4}{3}\pi\text{R}_{0}^{3}} = \frac{3}{4\pi\text{R}_{0}^{3}}$

The expression of the density is independent of mass number A.

1. Saturation/short range nature of nuclear forces.
2. We have

$\text{R} = \text{R}_{0}\text{A}^{1/3}$

$\therefore\text{ Density} \rho=\frac{\text{mA}}{\frac{4}{3}\pi\big(\text{R}_{0}\text{A}^{1/3}\big)^{3}}$

$ =\frac{\text{m}}{\frac{4}{3}\pi\text{R}_{0}^{3}}$

Hence $\rho$ is independent of A.
(Here m is the mass of the nucleus.)

Alternate Answer

For  $\alpha$ decay

$^{238}_{92}\text{U}\to^{234}_{90}\text{Th}+ ^{4}_{2}\text{He}+\text{Q}$

(or any other nuclear reaction)

$\text{For} ^{238}_{92}\text{U},\frac{\text{n}}{\text{p}} = \frac { 238 - 92}{92} = \frac{146}{92} =1.587$

$\text{For} ^{234}_{90}\text{Th},\frac{\text{n}}{\text{p}} = \frac { 234 - 90}{90} = 1.6$

$\therefore \frac{\text{n}}{\text{p}} $  increases during $\alpha$ decay.

For $\beta$ decay

$^{60}_{27}\text{Co}\to^{60}_{28}\text{Ni}+ ^{0}_{-1}\text{e}+ \text{Q}$

(or any other nuclear reaction)

$\text{For} ^{60}_{27}\text{Co},\frac{\text{n}}{\text{p}} = \frac{60 - 27 }{27} = 1.22$

$\text{For} ^{60}_{28}\text{Ni},\frac{\text{n}}{\text{p}} = \frac{60 - 28 }{28} = 1.14$

$\therefore \frac{\text{n}}{\text{p}} $ decreases during $\beta$ decay

Alternate Answer 

Full credit to be given if explained in terms of.

$^{\text{A}}_\text{Z}\text{X}\to^{\text{A-4}}_\text{Z-2}\text{Y}+ ^{4}_{2}\text{He}$

and

$^{\text{A}}_\text{Z}\text{X}\to^{\text{A}}_\text{Z+1}\text{Y}+ ^\circ_{1}\text{e} + {\overline{\text{V}}}\big(\text{V}\big)$

$\text{for }\alpha\text{ decay and }\beta\text{ decay}.$

Nuclear fission: Binding energy per nucleon is smaller for heavier nuclei (greater than 200) i.e., heavier nuclei are less stable. When a heavier nucleus splits into the lighter nuclei, they become stable, this happens in nuclear fission.

Nuclear fusion: The binding energy per nucleon is small for light nuclei (between 2 to 20), i.e., they are less stable. So when two light nuclei combine to form a heavier nucleus, the higher binding energy per nucleon of the latter results in the release of energy. This is what happens in a nuclear fusion.

2. $\frac{\text{R}}{\text{R}_0}=\frac{\text{N}}{\text{N}_0}=\Big(\frac{1}{2}\Big)^{\text{n}}$

$\frac{\text{R}}{\text{R}_0}=3.125\%=\frac{3.125}{100}=\frac{1}{32}=\Big(\frac{1}{2}\Big)^5$

$=\Big(\frac{1}{2}\Big)^\text{n}=\Big(\frac{1}{2}\Big)^5\Rightarrow\ \text{n}=5$

$\text{t}=\text{nT}_{\frac{1}{2}}=5\times10=50\text{ year}$