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Physics 2 Marks Questions

Nuclei · Rajasthan Board (RBSE) STD 12 Science (Rajasthan - English Medium). 15 questions.

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Question 12 Marks
The ratio of radii of two nuclei is $1: 2$. Write the ratio of their mass numbers.
Answer
Radius of nucleus $R=R_0 A^{1 / 3}$
or $A=\left(\frac{R}{R_0}\right)^3$
Here, $A =$ mass number, $R _0=$ constant
Therefore, $\frac{ A _1}{A_2}=\frac{ R _1^3}{ R _2^3}=\frac{1^3}{2^3}=\frac{1}{8}$
$A _1: A _2=1: 8$ Ans.
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Question 22 Marks
If the radius of the nucleus of $Al ^{27}$ is 3.6 Fermi , then what will be the radius of the $Fe ^{125}$ nucleus?
Answer
Since, $\quad R \propto(A)^{1 / 3}$
Therefore $\frac{ R _{ Fe }}{ R _{ Al }}=\left(\frac{ A _{ Fe }}{ A _{ A }}\right)^{1 / 3}=\left(\frac{125}{27}\right)^{1 / 3}=\frac{5}{3}$
$\therefore$ $R _{ Fe }=\frac{5}{3} \times R _{ Al }=\frac{5}{3} \times 3.6 Fermi$
$R _{ Fe }=6.0 Fermi$
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Question 32 Marks
Which physical quantities are conserved in natural and artificial radioactivity?
Answer
(1) Their charge is conserved.
(2) The sum of mass and energy remains conserved.
(3) Angular momentum and linear momentum are conserved.
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Question 42 Marks
Define atomic mass unit (amu) and write its equivalent energy.
Answer
One twelfth part of the mass of an atom of ${ }_6 C ^{12}$ is called atomic mass unit.
That is, $\quad 1 amu =\frac{1}{12} \times\left(\right.$ Mass of 1 atom of $\left.\left.{ }_6 C ^{12}\right)\right)$
$1 amu=1.66 \times 10^{-27} kg$
Energy equivalent of $1 amu =931 MeV$
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Question 52 Marks
What is the per nucleon binding energy?
Answer
If the binding energy of a nucleus with mass number A is $E _{ b }$, then the per nucleon binding energy is
$\frac{E_b}{F}=\frac{\Delta m \times c^2}{A}$
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Question 62 Marks
The ratio of mass numbers of two nuclei is $1: 8$. What is the ratio of their nuclear radii?
Answer
According to the question,
$A_1: A_2=1: 8$
$\begin{array}{ll}\text { or } & \frac{ A _1}{A_2}=\frac{1}{8}\end{array}$
$\because$ $R = R _0 A^{1 / 3}$
$\therefore$ $\frac{ R _1}{ R _2}=\frac{ A _1^{1 / 3}}{A_2^{1 / 3}}=\left(\frac{ A _1}{A_2}\right)^1$
$\frac{ R _1}{ R _2}=\left(\frac{1}{8}\right)^{1 / 3}=\frac{1}{2}$ Ans.
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Question 72 Marks
Mention two specific characteristics of nuclear force.
Answer
Specific characteristics :
(i) Nuclear forces are always independent.
(ii) Nuclear force is the strongest force of nature.
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Question 82 Marks
Write any two characteristics of nuclear force.
Answer
(1) These are the strongest forces found in nature.
(2) The range of nuclear forces is very short. Their range is of the order of the radius of the nucleus $\left(10^{-15}\right.$ meters). That is, these forces are effective only up to a distance of $10^{-15}$ meters. If this does not happen, then these forces are more effective than the electromagnetic interaction in the formation of molecules from atoms.
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Question 92 Marks
What is meant by radioactive equilibrium?
Answer
Radioactive Equilibrium : If the activity of a radioactive substance does not change with time during the disintegration process, then this state is called radioactive equilibrium. In this condition, the net rate of disintegration of all the elements except the first and last element of the radioactive process becomes zero at some time, i.e.
$\lambda_{A} N_{A}=\lambda_{B} N_{B}=\lambda_{C} N_{C} \ldots \ldots=\text { constant. }$
This constant is also called the activity of the radioactive sample.
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Question 102 Marks
Which rules are followed in nuclear reactions?
Answer
The following rules are followed in nuclear reactions :
(i) Charge is conserved.
(ii) Linear and angular momentum are conserved.
(iii) The sum of mass and energy remains conserved.
(iv) Nucleon is conserved.
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Question 112 Marks
What is the meaning of binding energy per nucleon? The binding energy per nucleon of neutron $\left({ }_1 H ^2\right)$ and $\alpha$-particle $\left({ }_2 He ^4\right)$ is 1.25 and 7.2 MeV per nucleon respectively. Which nucleus is more stable?
Answer
The average energy required to separate a nucleon from a nucleus is called binding energy per nucleon. The binding energy per nucleon of $\alpha$-particle is higher than that of the deuteron. Hence the $\alpha$-particle is more stable.
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Question 122 Marks
Name the nucleus for which the binding energy per nucleon is maximum.
Answer
The total binding energy of any nucleus depends on the number of nucleons present and the binding energy per nucleon represents the stability of the atom.
Binding energy per nucleon for iron ( Fe )
$\overline{B}=\frac{\Delta E}{A}=\frac{492.8}{56}=8.8 MeV$
which is the maximum.
Therefore this nucleus is the most stable.
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Question 132 Marks
The fission properties of ${ }_{94}^{239} Pu$ are very similar to those of ${ }_{92}^{235} U$. The average energy released per fission is 180 MeV . How much energy, in MeV, is released if all the atoms in 1 kg of pure ${ }_{94}^{239} Pu$ undergo fission?
Answer
$239 g{ }_{94} Pu^{239} \text { has } 6.023 \times 10^{23} \text { nuclei. }$
Therefore, $1000 g_{94}^{239} Pu$ will contain
$=\frac{6.023 \times 10^{23} \times 1000}{239}$
$=2.52 \times 10^{24}$ nuclei
Energy released by fission of each nucleus
$=180 MeV$
Therefore, Total energy released
$=180 \times 2.52 \times 10^{24} MeV$
$=4.54 \times 10^{26} MeV$
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Question 142 Marks
Suppose, we think the fission of a ${ }_{26}^{56} Fe$nucleus into two equal fragments, ${ }_{13}^{28} Al$.Is this fission energetically possible? Argue by working out Q of the process.
Given : $m\left({ }_{26}^{56} Fe \right)=55.93494 u$
and $\quad m\left({ }_{13}^{28} Al \right)=27.98191 u$
Answer
$Q =\left[m\left({ }_{26}^{56} Fe \right)-2 m\left({ }_{13}^{28} Al \right)\right] c^2$
$=(55.93494 u -2 \times 27.98191 u ) c^2$
$=(55.93494-55.96382) u c^2$
$=-0.02888 u \times 931.5 \frac{ MeV }{11}$
$=-0.02888 \times 931.5 MeV$
$=-26.90 MeV$
Obviously, 26.9 MeV energy will have to be supplied externally for fission of ${ }_{26} Fe ^{56}$ into two identical elements ${ }_{13} Al ^{28}$. Therefore, in this way disintegration of ${ }_{26} Fe ^{56}$ is not possible.
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Question 152 Marks
Obtain the binding energy (in MeV) of a nitrogen nucleus $\left({ }_7^{14} N\right)$, given $m\left({ }_7^{14} N\right)=14.00307 u$.
Answer
Given : $\quad Z=7$
And$\quad$$A =14$
$A-Z=14-7=7$
Mass of 7 protons $=7 \times 1.007825 u$
$=7.054775 u$
$\begin{aligned} \text { Mass of } 7 \text { neutrons } =7 \times 1.008665 u \\ =7.060655 u\end{aligned}$
$\begin{aligned} \text { Binding energy }=(14.11543 u- 14.00307 u) \\ \times 931.5 MeV \end{aligned}$
$=(0.11236) \times 931.5 MeV$
$=104.66 MeV$
$\simeq 104.7 MeV$
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