4 Marks Questions

Question 14 Marks

What do you mean by nuclear fission ?
On disintegration of one atom of ${}^{235}U$ the amount of energy obtained is 200 MeV. The power obtained in a reactor is 1000 kilowatt. How many atoms are disintegrated per second in the reactor? What is the decay in mass per hour?

Answer

Nuclear Fission : "The process in which two or more very light nuclei moving at very high speed i.e., possessing very high energy $(\approx 0.1 MeV)$ are fused together to form a single nucleus is called nuclear fusion."
In this process mass of the product nucleus is less than the total mass of the nuclei to be fused and the lost mass is converted into energy. The fusion process is, however, not easy to carry out. Since the nuclei to be fused are positively charged, they would repel one another strongly. Hence, they must be brought very close together not only by high pressure but also with high kinetic energy. For this, a temperature of the order of $10^8 K$ is required. Such high temperature is available in the Sun and stars. On earth they may be produced by exploding a nuclear fission bomb. Since very high temperature is needed for the fusion of nuclei, the process is called a 'thermonuclear reaction', and the energy released is called 'thermonuclear energy'.
Solution of Numercial:
Since power of reactor is 1000 kilowatt
= 10⁶watt
⇒ Energy received per second from reactor E
$=10^6$ joule
Disintegration of one atom of ${ }^{235} U$ gives energy
$E ^{\prime}=200 MeV$
$\Rightarrow \quad E^{\prime}=200 \times 1.6 \times 10^{-13}$ joule
$=3.2 \times 10^{-11}$ joule
∴ Number of atoms disintegrated per second
$n=\frac{E}{E^{\prime}}=\frac{10^6 \text { joule }}{3.2 \times 10^{-11} \text { joule }}$
$\Rightarrow \quad=3.125 \times 10^{16}$ atoms
Energy received per hour $E ^{\prime \prime}= E \times 60 \times 60 sec$
$=\left(10^6 J / s \right) \times 3600 s=36 \times 10^8$ joule
$\because \quad E ^{\prime \prime}=\Delta m \times c^2$
⇒ Decay in mass
$\Delta m=\frac{E^{\prime \prime}}{c^2}=\frac{36 \times 10^8 \text { joule }}{\left(3 \times 10^8 m / s \right)^2}= 4 \times 1 0 ^{-8} ~ k g$

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Question 24 Marks

Explain the difference between nuclear fission and nuclear fusion.

Answer

Difference between Nuclear fission and Nuclear Fusion.

Nuclear Fission	Nuclear Fusion
1. When the nucleus of an atom splits into lighter nuclei through a nuclear reaction, the process is termed nuclear fission.	Nuclear fusion is a reaction through which two or more light nuclei collide with each other to form a heavier nucleus.
2. When each atom splits, a tremendous amount of energy is released.	The energy released during nuclear fusion is several times greater than the energy released during nuclear fission.
3. Fission reactions do not occur in nature naturally.	Fusion reactions occur in stars and the sun.
4. Comparatively, less energy is needed to split an atom in a fission reaction.	High energy is needed to fuse two or more atoms together in a fusion reaction.

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Question 34 Marks

Write difference between nuclear fission and radioactivity i.e. radioactive decay.

Answer

Difference between Nuclear fission and Radioactive decay : Although these two processes are nuclear phenomenon, yet they are quite different from each other in following respects :
(i) Radioactive decay is a spontaneous phenomenon whereas fission is not. In fission, heavy nuclei are bombarded by neutrons.
(ii) In radioactive decay, $\alpha$ - and $\beta$ - particles are emitted from the nucleus and energy is obtained in the form of $\gamma$ - rays which is not very large. In nuclear fission, a heavy nucleus is broken into two nearly equal lighter nuclei and very huge energy is liberated.
(iii) In radioactive decay, the atomic number can change by 1 or 2 and the mass number can change by 0 or 1. In nuclear fission, both atomic number and mass number are almost equally distributed.
(iv) The rate of radioactive decay cannot by controlled in any way, but the rate of nuclear fission can be controlled.

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Question 44 Marks

What is nuclear force? Write its properties. What do you mean by stability of a nucleus ?

Answer

Nuclear Forces : A number of nucleons are present in a very small volume $(\approx10^{-44}m^{3})$ of the nucleus. As each neucleon has a finite mass $\left(\approx 1.6 \times 10^{-27} kg\right)$ and each proton has a finite charge $\left(\approx 1.6 \times 10^{-19} C \right)$, so the nucleons experience both the gravitational and electrostatic forces. The gravitational force which is attractive in nature is very weak because of the very small value of mass of each nucleon and very small value of $G(=6.67\times10^{-11}Nm^{2}/kg^{2})$. Due to very small distance between the protons the electrostatic force of repulsion between protons is very large which is $10^{36}$ times greater than gravitational force between them. Hence it is obvious that unless some other strong attractive force acts between the nucleons the repulsive electrostatic force between the protons would keep them faisly away from each other, resulting in the disruption of nuclei. "This new force which keeps the nucleons within the nucleus bound together is called nuclear force." The following are the well known facts about these forces.
Properties of nuclear forces :
(i) These are attractive in nature.
(ii) These are the strongest forces out of all the forces known till now.
(iii) These are not electrostatic forces.
(iv) These are not gravitational forces.
(v) These are extremely short range forces effective within nuclear diameter $\approx10^{-15}$ m.
(vi) These forces are independent of charge i.e. attractive force between neutron-neutron, proton-proton and proton-neutron is the same.
(vii) These are non-central forces i.e., the nuclear force between two nucleons does not act along the line joining their centres.
(viii) These forces have saturation character i.e. a nucleon interacts only with its neighbouring nucleon.
Stability of Nucleus : The stability of a nucleus is decided by the 'relative' number of protons and neutrons present in the nucleus. The, nuclei of lighter elements (except hydrogen) have equal, or nearly equal, number of protons and neutrons. In these nuclei, the very strong attractive nuclear forces completely overcome the electrical repulsive forces acting between the protons. Hence the lighter elements remain stable. As we move towards the heavier elements, the number of protons and neutrons both increases in their nuclei, but the number of neutrons increases more rapidly compared to the number of protons. In iron the number of neutrons is nearly 20% more than the number of protons, while in uranium neutrons are nearly 50% more. Since the (electrical) repulsive forces act between every pair of protons, whereas the nuclear attractive forces (being short-range) are active only between nucleons extremely close to each other, therefore, the total electrical repulsive force rises more rapidly compared to the nuclear attractive force. As a result, the stability of nucleus goes on decreasing. As the number of protons (also of neutrons) increases, the stability of the nucleus decreases more and more. This is the reason that all elements heavier than lead (wherein number of protons is more than 82) are unstable. They continue radioactive emission and are converted into lighter elements. This is why only very heavy elements are radioactive.

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