Physics 4 Marks Questions

1. Here we take A = 8 × 108 dis./sec

1. ​​​​​$\text{ln}\Big(\frac{\text{A}_1}{\text{A}_{0_{1}}}\Big)=\text{ln}\Big(\frac{11.79}{8}\Big)=0.389$

2. $\text{ln}\Big(\frac{\text{A}_2}{\text{A}_{0_{2}}}\Big)=\text{ln}\Big(\frac{9.1680}{8}\Big)=0.1362$

3. $\text{ln}\Big(\frac{\text{A}_3}{\text{A}_{0_{3}}}\Big)=\text{ln}\Big(\frac{7.4492}{8}\Big)=-0.072$

4. $\text{ln}\Big(\frac{\text{A}_4}{\text{A}_{0_{4}}}\Big)=\text{ln}\Big(\frac{6.2684}{8}\Big)=-0.244$

5. $\text{ln}\Big(\frac{5.4115}{8}\Big)=-0.391$

6. $\text{ln}\Big(\frac{3.0828}{8}\Big)=-0.954$

7. $\text{ln}\Big(\frac{1.8899}{8}\Big)=-1.443$

8. $\text{ln}\Big(\frac{1.167}{8}\Big)=-1.93$

9. $\text{ln}\Big(\frac{0.7212}{8}\Big)=-2.406$

2. ​​​​​​​The half life of ¹¹⁰Ag from this part of the plot is 24.4s.
3. Half life of ¹¹⁰Ag = 24.4s.

$\therefore\text{decay constant}\lambda=\frac{0.693}{24.4}=0.0284\Rightarrow\text{t}=50\text{sec,}$

The activity $\text{A = A}_{0}\text{e}^{-\lambda\text{t}}=8\times10^8\times\text{e}^{-0.0284\times50}=1.93\times10^8$

5. The half life period of ¹⁰⁸Ag from the graph is 144s.

1. (d) 2.4 × 10¹⁷kg m^-3

Explanation:

Here R = 3 × 10^-15m

Nuclear mass = 16 amu = 16 × 1.66 × 10^-27kg

$\rho_\text{nu}=\frac{\text{Nuclear mass}}{\text{Nuclear volume}}=\frac{16\times1.66\times10^{27}}{\frac{4}{3}\pi(3\times10^{-15})^3}$

= 2.359 × 10¹⁷kg m^-3 = 2.4 × 10¹⁷kg m^-3

2. (a) 1.2 × 10¹⁷kg m^-3

Explanation:

$\text{Density}\ \rho=\frac{3\text{m}}{4\pi\text{R}_0^3}=\frac{3\times1.007825\times1.66\times10^{-27}}{4\times\frac{22}{7}\times(1.1\times10^{-15})^3}$

= 2.98 × 10¹⁷kg m^-3

3. (d) A constant.

Explanation:

$\text{Density}=\frac{\text{Mass}}{\text{Volume}}=\frac{\text{Am}_\text{p}}{\frac{4}{3}\pi\big(\text{R}_0\text{A}\frac{1}{3}\big)}$

$=\frac{\text{m}_\text{p}}{\frac{4}{3}\pi\text{R}_0^3}$

Where m_p = 1.6 × 10^-27kg³ = 2.3 × 10¹⁷kg m-3, which is a constant.

4. (c) 2.9 × 10¹⁷kg m^-3

Explanation:

Given, mass of m^Fe = 55.85amu

= 55.85 × 1.66 × 10^-21kg = 9.27 × 10^-26kg

Nuclear radius $=\text{R}_0\text{A}\frac{1}{3}=1.1\times10^{-15}\times(56)\frac{1}{3}\text{m}$ 

[$\therefore$ A = 56]

$\rho_\text{nu}=\frac{\text{Nuclear mass}}{\text{Nuclear volume}}=\frac{\text{m}_\text{fe}}{\frac{4}{3}\pi\text{R}^3}$

$=\frac{9.27\times10^{26}}{\frac{4\pi}{3}\times\big(1.1\times10^{-15}\big)^3\times56}=2.9\times10^{17}\text{Kg}\text{ m}^{-3}$

5. (d) 6fm

Explanation:

Here A₁. = 27, A₂ = 125, R₁ = 3.6fm

$\text{As},\frac{\text{R}_2}{\text{R}_1}=\bigg(\frac{\text{A}_2}{\text{A}_1}\bigg)^\frac{1}{3}=\bigg(\frac{125}{27}\bigg)^\frac{1}{3}=\frac{5}{3}$

$\therefore\text{R}_2=\frac{5}{3}\text{R}_1=\frac{5}{3}\times3.6=6\text{fm}$

1. (a) 1

Explanation

As nearly 99.9% mass of atom is in nucleus 

$\therefore\frac{\text{Mass of nucleus}}{\text{Mass of atom}}=\frac{99.9}{100}=0.99\approx1$

2. (a) 1 : 2 : 3

Explanation:

Since, the nuclei of deuterium and tritium are isotopes of hydrogen, they must contain only one proton each. But the masses of the nuclei of hydrogen, deuterium and tritium are in the ration of 1 : 2 : 3, because of presence of neutral matter in deuterium and tritium nuclei.

3. (c) Isotones
4. (a) A straight line

Explanation:

$\text{R}=\text{R}_0\text{A}\frac{1}{3}$

$\text{log}\text{ R}=\text{log}\text{ R}_0+\frac{1}{3}\text{log}\text{A}$

On comparing the above equation of straight tine; y = mx + c. So, the graph between log A and log R is a straight line also.

5. (a) 1.23

Explanation

Here, A1 = 197 and A₂ = 107

$\therefore\frac{\text{R}_1}{\text{R}_2}=\Big(\frac{\text{A}_1}{\text{A}_2}\Big)^\frac{1}{3}=1.225=1.23$

Carbon dating: Radiocarbon dating (also referred to as carbon dating or carbon-14 dating) is a method tor determining the age of an object containing organic material by using the properties of radiocarbon -14(¹⁴C), a radioactive isotope of carbon.

Radiocarbon, or carbon 14, is an isotope of the element carbon that is unstable and weakly radioactive. The stable isotopes are carbon 12 and carbon 13.

Carbon 14 is continually being formed in the upper atmosphere by the effect of cosmic ray neutrons on nitrogen 14 atoms. It is rapidly oxidized in air to form carbon dioxide and enters the global carbon cycle.

Plants and animals assimilate carbon 14 from carbon dioxide throughout their lifetimes. When they die, they stop exchanging carbon with the biosphere and their carbon 14 content then starts to decrease at a rate determined by the law of radioactive decay.

Radiocarbon dating is essentially a method designed to measure residual radioactivity. By knowing how much carbon 14 is left in a sample, the age of the organism when it died can be known. It must be noted though that radiocarbon dating results indicate when the organism was alive but not when a material from that organism was used.

It is given that

The ¹⁴C activity of a piece of wood from the ruins of an ancint building,

R = 12dis/min per g

The ¹⁴C activity of the living wood, R₀ = 16dis/min per g

Half-life of ¹⁴C $\text{T}_\frac{1}{2}=5760\text{yr}$

Let t be the span of the tree.

According to radioactive decay law,

$\text{R}=\text{R}_0\text{e}^{-\lambda\text{t}\text{ or }}\frac{\text{R}}{\text{R}_0}=\text{e}^{-\lambda\text{t}}\text{ or }\text{e}^{\lambda\text{t}}=\frac{\text{R}_0}{\text{R}}$

Taking log on both sides,

$\lambda\text{t}\log_\text{e}\text{e}=\log_\text{e}\frac{\text{R}_0}{\text{R}}\Rightarrow\ \lambda\text{t}=\Big(\log_{10}\frac{16}{12}\Big)\times2.303$

$\text{t}=\frac{2.303(\log4-\log3)}{\lambda}=\frac{2.303(\log4-\log3)\text{T}_\frac{1}{2}}{0.6931}$

$=\frac{2.303(0.6020-4.771)\times5760}{0.6931}=2391.20\text{yr}$

1. (a) Total energy, but not kinetic energy.

Explanation:

As just shown, energy conservation allows us to calculate the third particle's total energy. But we do not know what percentage of that total is mass energy. 

2. (d) It is less than or equal to 0.40 MeV, but we cannot be more precise.

Explanation:

According to the passage, subatomic reactions do not conserve mass. So, we cannot find the third particle's mass by setting $\text{m}_\text{neutron}$equal to $\text{m}_\text{proton}+\text{m}_\text{electron}+\text{E}_\text{thrid particle}$

The neutron has energy 940.97 MeV. The proton has energy 939.67 MeV + 0.01 MeV = 939.69 MeV. The electron has energy 0.51 MeV + 0.39 MeV = 0.90 MeV. Therefore, the third particle has energy.

$\text{E}_\text{third particle}=\text{E}_\text{neutron}-\text{E}_\text{proton}-\text{E}_\text{electron}$

= 940.97 - 939.67 - 0.90 = 0.40 MeV 

We just found the third particle's total energy, the sum of its mass energy and kinetic energy. Without more information, we cannot figure out how much of that energy is mass energy.

3. (b) Yes, but only if the proton has potential energy (due to interactions with other particles). 
4. (a) 2 × 10^-5kg

Explanation:

Here, P = 500MW = 5 × 10⁸W,

t = 1h = 3600s

Energy produced, E = P × t = 5 × 10⁸ × 3600 = 18 × 10¹¹J

$\text{As}\ \text{E}=\Delta\text{mc}^2$

$\therefore\Delta\text{m}=\frac{\text{E}}{\text{c}^2}=\frac{18\times10^{11}}{(3\times10^8)^2}$

$=\frac{18\times10^{11}}{(3\times10^8)^{16}}=2\times10^{-5}\text{kg}$

5. (a) 9 × 10¹³J

Explanation:

Using, E = mc²

Here, m = 1g = 1 × 10^-3kg, c = 3 × 10⁸m s^-1

$\therefore$ E = 10^-3 × 9 × 10¹⁶ = 9 × 10¹³J

1. (a) 3.125 × 10¹³

Explanation:

Let the number of fissions per second be n. Energy released per second

= n × 200MeV = n × 200 × 1.6 × 10^-13J

Energy required per second = power × time

= 1kW × Is = 1000J

n × 200 × 1.6 × 10^-13 = 1000 or 

$\text{n}=\frac{1000}{3.2\times10^{-11}}$

$=\frac{10}{3.2}\times10^{13}=3.125\times10^{13}$

2. (a) More mass per nucleon than either of the two fragments.
3. (b) 9 × 10¹³J

Explanation:

As only 0.1% of the original mass is converted into energy, hence out of 1kg mass 1g is converted into energy.

Energy released during fission $\text{E}=\triangle\text{mc}^2$

= 1g × (3 × 10⁸ m s^-1)² = 10^-3 × 9 × 10¹⁶J = 9 × 10¹³J

4. (a) K = 1
5. (a) Nuclear fission

1. (d) All of these.

Explanation:

All options are basic properties of nuclear forces. So, all options are correct.

2. (d) 1.4 × 10^-15m

Explanation:

The nuclear force is of short range and the range of nuclear force is the order of 1.4 × 10^-15m. Now, volume $\infty\text{R}^3\infty\text{A}$

3. (d) Strong nuclear force.
4. (a) F_n << F_e

Explanation:

Nuclear force is much stronger than the electrostatic force inside the nucleus i.e., at distances of the order of fermi. At 40 A, nuclear force is ineffective and only electrostatic force of repulsion is present. This is very high at this distance because nuclear force is not acting now and the gravitational force is very feeble. F_nuclear << F_{elcctrostatic} in this case.

5. (a) Nuclear forces

1. (b) Liquid drop model
2. (b) Elastic damping/ scattering

Explanation:

Fast neutrons are slowed down by elastic scattering with light nuclei as each collision takes away nearly 50% of energy.

3. (d) Both I and II

Explanation:

Reactions I and II represent fission of uranium isotope $_{92}^{235}\text{U}$ when bombarded with neutrons that breaks it into two intermediate mass nuclear fragments. However, reaction III represents two deuterons fuses together to from the light isotope of helium.

4. (c) 2.5 and 2 MeV

Explanation:

On an average 2.5 neutrons are released per fission of the uranium atom. The energy of the neutron released per fission of the uranium atom is 2 MeV.

5. (a) Less than I

Explanation:

In fission process, when a parent nucleus breaks into daughter products, then some mass is lost in the form of energy. Thus, mass of fission products < mass of parent nucleus.

$\Rightarrow\frac{\text{Mass of fission products}}{\text{Mass of parent nucleus}}<1$