3 Marks Question

Question 13 Marks

Explain the process of thermonuclear fusion with example.

Answer

→When two light nuclei fuse to form a larger nucleus, energy is released, because binding energy increases during the process.
→Some examples of such energy liberating nuclear fusion reactions are :
$\begin{array}{l}
{ }_1 H ^1+{ }_1 H ^1 \rightarrow{ }_1 H ^2+e^{+}+v+0.42 MeV \\
{ }_1 H ^2+{ }_1 H ^2 \rightarrow{ }_2 He ^3+n+3.27 MeV \\
{ }_1 H ^2+{ }_1 H ^2 \rightarrow{ }_1 H ^3+{ }_1 H ^1+4.03 MeV
\end{array}$
→In the first reaction, two protons combine to form a deuteron and a positron with a release of 0.42 MeV energy.
→In the second reaction, two deuterons combine to form the isotope of helium ${ }_2 He ^3$.
→In third reaction two deuterons combine to form a tritium and a proton. 4.03 MeV energy is released during this process.
→For a fusion to take place, the two nuclei must come close enough so that nuclear force is able to affect them.
→This force must be strong enough to overcome the repulsive barrier between two positively charged nuclei.
→The height of the barrier depends on the charges and radii of the two interacting nuclei.
→For example :
The barrier height for two protons is $\sim 400 keV$. The temperature required for a proton to overcome this barrier is T .
$\begin{aligned}
\therefore \quad \frac{3}{2} k T & =400 keV \\
T & =\frac{2 \times 400 \times 10^3 \times 1.6 \times 10^{-19}}{3 \times 1.38 \times 10^{-23}} \\
T & =3 \times 10^9 K
\end{aligned}$
→When fusion is achieved by raising the temperature of the system so that particles have enough kinetic energy to overcome the Coulomb repulsive barrier, it is called thermo - nuclear fusion.

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Question 23 Marks

Explain isotopes, isobars, isotones and isomers by giving examples.

Answer

→ (i) Isotopes :
→The atoms which have atomic number Z same but atomic mass number A different, then such type of atoms are called the isotopes of each other.
→For example :
Isotopes of hydrogen are ${ }_1 H ^1,{ }_1 H ^2,{ }_1 H ^3$
•${ }_1 H ^1$ - there is one proton but having no neutron.
• ${ }_1 H ^2$ - there is one proton one neutron.
• ${ }_1 H ^3$ - there is one proton and two neutrons.
→Isotopes of carbon are ${ }_6 C ^{12},{ }_6 C ^{13},{ }_6 C ^{14}$
→Isotopes of uranium are ${ }_{92} U ^{233},{ }_{92} C ^{235},{ }_{92} C ^{238}$
→ (ii) Isobar :
→Atoms having same atomic mass number A , but different atomic number Z are called the isobars of each other.
→ For example :
${ }_1 H ^3$ and ${ }_2 He ^3$
${ }_{82} Pb^{214}$ and ${ }_{83} Bi ^{214}$
→ (iii) Isotone :
The atoms for which the neutron number N is the same but atomic number Z and mass number A are different are called isotones to each other.
→ For example :
${ }_{80} Hg ^{198} \text { and }{ }_{79} Au ^{197}$
→ (iv) Isomer :
→The atoms for which the atomic number Z and mass number A are same but their radioactive properties are different are called isomers of each other.
→For example :
${ }_{35} Br ^{80}$ contains a pair of isomers.

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Question 33 Marks

What are isotopes? Explain with an example how the average mass can be obtained from the relative proportions of different isotopes of the same element.

Answer

→The atoms for which atomic number Z is same but atomic mass number A is different then such type of atoms are called the isotopes of each other.
→Almost every element is a mixture of many isotopes. The relative abundance of different isotopes differs from element to element.
→For example :
(i) Chlorine has two isotopes having masses $34.98 u$ and $36.98 u$. The relative abundances of these isotopes are $75.4 \%$ and $24.6 \%$.
→Average mass of Chlorine
$\begin{array}{l}
=\frac{75.4 \times 34.98+24.6 \times 36.98}{100} \\
=35.47 u
\end{array}$
→This mass is almost equal to the atomic mass of chlorine.
(ii) Hydrogen also has three isotopes. Their masses are $1.0078 u, 2.0141 u$ and $3.0160 u$.
→The nucleus of the lightest atom of hydrogen has a relative abundance of $99.985 \%$, is called proton but there is no neutron.
→The other two isotopes of hydrogen are called deuterium (mass $=2.0141 u$ ) and tritium (mass $=3.0160 u$ ).
→Tritium nuclei being unstable, do not occur naturally and are produced artificially in laboratories.
→The relative abundance of deuterium is so small (relative abundance of hydrogen is $99.985 \%$ ) that the masses of deuterium and tritium are neglected when calculating the average mass of hydrogen.
→Hydrogen $\left({ }_1 H ^1\right)$ nucleus has only one proton and do not have neutrons.
→Mass of proton (mass of ${ }_1 H ^{ 1}$ )
$\begin{array}{l}
=\frac{1.0078 \times 99.985}{100} \\
=1.00727 u \\
=1.00727 \times 1.660539 \times 10^{-27} kg \\
=1.67262 \times 10^{-27} kg
\end{array}$
→This value is equal to the value obtained by subtracting the mass of an electron from the mass of a hydrogen atom.

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Question 43 Marks

Define the different terms used for the composition of a nucleus.

Answer

• The Composition of nucleus :
(i) Atomic number $( Z )$ :
→The number of electrons in the neutral atom is called the atomic number.
OR
→The number of protons in the neutral atom is called the atomic number.
(ii) Mass number (A) :
→The sum of the number of protons and neutrons in the nucleus of the atom is called the atomic mass number A .
→Both protons and neutrons are present in the nucleus hence it is known by the common name of nucleon.
→The total number of nucleons in the nucleus of an atom is called atomic mass number A .
$\therefore A = Z + N$
(iii) Neutron number ( N $)$ :
→The number of neutrons present in the nucleus of an atom is called the neutron number N .
(iv) Nuclide :
→The type of nucleus is called nuclide and is represented by symbol ${ }_Z X^A$.
→Where X is the chemical symbol of element, $Z =$ atomic number and A is atomic mass number.
For example :
Nuclide of gold ${ }_{79} A u^{197}$
in which 197 nucleons, 79 protons and 118 neutrons are there.

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Question 53 Marks

Explain the phenomenon of nuclear fission.

Answer

→When a heavy nucleus is bombarded with a neutron, then first neutron is absorbed. This nucleus is in a highly excited state. As a result, it splits into two lighter nuclei of approximately equal mass to become stable.
→Neutron is chargeless so it does not have to face coulomb forces. So neutron is a best projectile.
→When a neutron is bombarded on the nucleus of uranium its nucleus breaks into two almost equal parts. Its nuclear reaction is as below :
$\begin{array}{l}
{ }_{92}^{235} U +{ }_0^1 n \rightarrow{ }_{92}^{236} U \rightarrow{ }_{56}^{144} B a+{ }_{36}^{89} K r+3\left({ }_0^1 n\right)+ Q \\
{ }_{92}^{235} U +{ }_0^1 n \rightarrow{ }_{92}^{236} U \rightarrow{ }_{51}^{133} S b+{ }_{41}^{99} N b+4\left({ }_0^1 n\right)+ Q \\
{ }_{92}^{235} U +{ }_0^1 n \rightarrow{ }_{92}^{236} U \rightarrow{ }_{54}^{140} Xe +{ }_{38}^{94} S r+2\left({ }_0^1 n\right)+ Q
\end{array}$
→The fission fragments are radioactive and by successive emmision of $\beta$-particles results in the stable nuclei.
→During the fission process of uranium the energy released per fission is almost 200 MeV .
→Suppose a nucleus with mass number $A =240$ breaks into two fragments each of $A =120$.
→Binding energy per nucleon for a nucleus with $A =240$ is 7.6 MeV and for a nucleus with $A =120$ is 8.5 MeV .
→Gain in binding energy per nucleon
$\begin{array}{l}
=8.5-7.6 \\
=0.9 MeV
\end{array}$
→Total gain in binding energy
$\begin{array}{l}
=0.9 \times 240 \\
=216 MeV .
\end{array}$
→The disintegration energy in fission events first appears as the kinetic energy of the fragments and neutrons. Eventually it is transferred to the surrounding matter appearing as heat.
→In a nuclear reactor this process takes place in a controlled manner whereas in an atomic bomb this process takes place in an uncontrolled manner.

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Question 63 Marks

Draw the graph of binding energy per nucleon against atomic mass number and draw conclusion of the graph.

Answer

→(i) The force is attractive and sufficiently strong to produce a binding energy of a few MeV per nucleon.
(ii) The constancy of the binding energy in the range $30< A <170$ is a consequence of the fact that the nuclear force is short ranged.
→Consider a particular nucleon inside a sufficiently large nucleus. This nucleon can interact with another nucleon within a distance of the duration of the nuclear force.
→If any other nucleon is at a distance more than the range of the nuclear force from the particular nucleon, so there is no interaction between these two nucleus. (There is no force between two nucleons)
→Suppose a nucleon can have a maximum of $p$ neighbours within the range of the nuclear force. All these nucleons exert a force on the considered nucleon. Due to which the considered nucleon has binding energy. This binding energy would be proportional to $p$. Let the binding energy of the nucleus be $p k$, where $k$ is a constant having the dimension of energy.
→If we increase A by adding nucleons they will not change the binding energy of a nucleon inside.
→Since most of the nucleons in a large nucleus reside inside it and not on the surface, the change in binding energy per nucleon would be small, which can be ignored. The binding energy per nucleon is constant and is approximately equal to $p k$.
→The property that a given nucleon influences only nucleons close to it is also referred to as saturation property of the nuclear force.
(iii) A very heavy nucleus say $A =240$, has lower binding energy per nucleon compared to that of a nucleus with $A =120$.
→ Thus, if a nucleus $A =240$ breaks into two $A =120$ nuclei, so the value of $E _{b n i}$ increases. This implies energy would be released in the process. This process is called Nuclear Fission. This process takes place in a controlled manner in a nuclear reactor.
→"Consider two very light nuclei $(A \leq 10)$ joining to form a heavier nucleus, the binding energy per nucleon increases. Energy is also released during this process. This process is called nuclear fusion". Energy is released from the sun due to the thermal nuclear fusion that takes place in the Sun.

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Physics 3 Marks Question

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