WBCHSE Class 12 Physics Atomic Nucleus Question And Answers

WBCHSE Class 12 Physics Atomic Nucleus Question And Answers

Question 1. Energy evolved during nuclear fission can be used for the welfare of mankind’ as discussed briefly.
Answer:

A thermal neutron is used to disintegrate U-235 resulting in the emission of either 2 or 3 neutrons which in turn disintegrate U-235 again in the next stop. So, In a very short time, the number of disintegrated nuclei increases in multiples and the reaction becomes uncontrolled. The energy evolved due to this uncontrolled chain reaction leading to an explosion.

However, If the chain reaction is maintained in a controlled manner by using only one neutron to bombard the U-235 in each step, the energy produced can be utilized for the welfare of mankind. The reaction will be under control. This will help to evolve energy at the same rate as the rate of the reaction. Usually, nuclear reactors are used for this purpose. Boron is used in these reactions to absorb the excess neutrons

Question 2.  Electromagnetic waves are emitted from an atom in an excited state and γ -rays are emitted during radioactive disintegration. What are the similarities and dissimilarities between them
Answer:

Similarities:

  1. Both are electromagnetic waves and travel with the same velocity.
  2. Both are not deflected by electric and magnetic fields.

Dissimilarities:

Atomic Nucleus Electromagnetic Waves And Gama Rays

Question 3. Write two characteristic features of the nuclear force that distinguish it from the Coulomb force
Answer:

Nuclear force is charge-independent. If the distance remains the same, there is no difference between the proton-proton, proton-neutron, and neutron-neutron force. On the other hand, coulomb force acts between charged particles only

The nuclear force is a short-range force; it is a strong attractive force within a distance of about 10-1’1 m and becomes zero outside that distance. But, Coulomb force has an infinite range

Question 4. A mixture consists of two radioactive materials Ay and decayed? A2 with half-lives of the 20s and 10s respectively, initially the mixture has 40 g of A1 and 160 g of A2. After what time the amount of the two In the mixture will become equal?
Answer:

Let after t s, quantity of A1 and A2 are equal.

We know N = \(N_0\left(\frac{1}{2}\right)^{t / T}\)

where T = Half life

For \(A_1, N_1=N_{01}\left(\frac{1}{2}\right)^{t / 20}\) and For, \(A_2, N_2=N_{02}\left(\frac{1}{2}\right)^{t / 10}\)

N1 = N2

Or, \(40 \times\left(\frac{1}{2}\right)^{t / 20}=160\left(\frac{1}{2}\right)^{t / 10}\)

Or, \(2^{2-\frac{t}{20}}=2^{4-\frac{t}{10}}\)

Or, \(2-\frac{t}{20}=4-\frac{t}{10}\)

Or, t = 40 s

Question 5. The half-life of a radioactive nucleus is 50 days. What is the time interval (t2– f1) between the time t2 when 2/3 of it has decayed and the time t1 when 1/3  of it has decayed?
Answer:

From the law of nuclear decay

N = N0e-λt

If \(\frac{1}{3}\) rd of nucleus decays in time \(t_1, \frac{1}{3} N_0=N_0 e^{-\lambda t_1}\)

And if\(\frac{2}{3}\) rd of nucleus decays in time \(\)

⇒ \(\frac{\frac{1}{3} N_0}{\frac{2}{3} N_0}=\frac{e^{-\lambda t_1}}{e^{-\lambda t_2}} \quad \text { or, } \frac{1}{2}=e^{-\lambda\left(t_2-t_1\right)}\)

Or, \(t_2, \frac{2}{3} N_0=N_0 e^{-\lambda t_2}\)

Putting \(\lambda=\frac{\ln 2}{T_{1 / 2}}\) and solving , we get

⇒ \(t_2-t_1=T_{1 / 2}\)

= 50 days

Question 6. Obtain the binding energy of the nuclei 56Fe26  and 209Bi83 in units of meV from the following data mH = 1.007825 u, mn = 1.008665 unchanged,

⇒  \(m\left({ }_{26}^{56} \mathrm{Fe}\right)=55.934939 \mathrm{u} m\left({ }_{83}^{209} \mathrm{Bi}\right)=208.980388 \mathrm{u}\) Which nucleus has greater binding energy per nucleon?
Answer:

For 56Fe26 mass, the defect

Δm = ZmH + (A-Z)mn-m(2gFe)

Δm = 26 × 1.007825 + (56-26) ×  1.008665-55.934939

∴ Binding energy =Δ m × 931.2 MeV = 492.2 MeV

∴ Binding energy per nucleon = \(\frac{492.2}{56}\)

= 8.79 MeV

For \({ }_{83}^{209} \mathrm{Bi}\) , mass defect

\(\Delta m=83 \times 1.007825+(209-83)\)\(\times 1.008665-208.980388\)

∴ Binding cnergy= 1.760877 X 931.5 MeV = 1640.25 MeV

∴ A Binding energy per nucleon = \(\frac{1640.26}{209}\)

= 7.85 MeV

Question 7. A given coin has a mass of 3.0 g. Calculate the nuclear energy that would be required to separate all the neutrons and protons from each other. For simplicity assume that the coin is entirely made of 63Cu29 atoms (of mass 62.92960 u ). The masses of protons and neutrons are 1.00783 u and 1.00867 u respectively.
Answer:

Binding energy = \(Z m_p+(A-Z) m_n-m\left({ }_{29}^{63} \mathrm{Cu}\right)\) ×  931.5

= 0.59225 ×  931.5 MeV

Number of atoms in 3 g of Cu = \(\frac{6.023 \times 10^{23} \times 3}{63}\)

∴ The binding energy of 3 g of Cu

= \(\frac{0.59225 \times 931.5 \times 6.023 \times 10^{23} \times 3}{63}\)

= 1.58 × 1025  MeV

Question 8. The half-life is 28 years. What is the disintegration rate of 15 mg of this isotope?
Answer:

Number of atoms in 15 mg ofÿgSr

= \(\frac{6.023 \times 10^{23} \times 15 \times 10^{-3}}{90}\)

Disintegration constant,

= \(\frac{0.693}{T}=\frac{0.693}{28 \times 365 \times 24 \times 60 \times 60} \mathrm{~s}^{-1}\)

Rate of disintegration

= \(\frac{d N}{d t}=\lambda N\)

= \(\frac{0.693}{28 \times 365 \times 24 \times 60 \times 60}\)\(\times \frac{6.023 \times 10^{23} \times 15 \times 10^{-3}}{90}\)

= 7.87 × 10 Bq

Question 9. Find Q -value and kinetic energy of the emitted α -particle a -decay of 

  1. In the α -decay  226Ra88 ‘
  2. 226Rn86  

Given , \(m\left({ }_{86}^{226} \mathrm{Ra}\right)=226.02540 \mathrm{u}, m\left({ }_{86}^{222} \mathrm{Rn}\right)=222.01750 \mathrm{u}\)

⇒ \(m\left({ }_{86}^{220} \mathrm{Rn}\right)=220.01137 \mathrm{u}, m\left({ }_{84}^{216} \mathrm{Po}\right)=216.00189 \mathrm{u}\)

⇒ \(m\left({ }_2^4 \mathrm{He}\right)=4.00260 \mathrm{u}\)

Answer:

1. \({ }_{86}^{226} \mathrm{Ra} \rightarrow{ }_{86}^{222} \mathrm{Rn}+{ }_2^4 \mathrm{He}\)

Q – Value \(=m\left({ }_{88}^{226} \mathrm{Ra}\right)-\left[m\left({ }_{86}^{222} \mathrm{Rn}\right)+m\left({ }_2^4 \mathrm{He}\right)\right]\)

226,02540- (1222.0175 + 4.00260)

= 0.0053 u = 0.0053 × 931.2 MeV

= 4.93 MeV

The kinetic energy of α -particle

= \(\left(\frac{A-4}{A}\right) Q=\frac{222-4}{222} \times 4.93\)

= 4.84 MeV

2. Similarly for 220 Rn 86

Q -value = 6.41 MeV and kinetic energy of α  – particle

= 6.29 MeV

Question 10. The radionuclide 11C6 decays accroding to \({ }_6^{11}\mathrm{C}\rightarrow{}_5^{11}\mathrm{~B}+\mathrm{e}^{+}+\nu\) . T1/2 = 20.3 min. The maximum energy of the emitted positron is 0.960 MeV. Calculate Q and compare it with the maximum energy of the positron emitted.Given M(11C6) = 11. 01143 u, m(11B5) = 11.009305 u , me = 0.000548 u
Answer:

Q = \(\left[m\left({ }_6^{11} \mathrm{C}\right)-6 m_e\right]-\left[m\left({ }_5^{11} \mathrm{~B}\right)-5 m_e+m_e\right]\)

(mN stands for the nuclear mass of the element or particle
= atomic mass – a mass of extranuclear electrons).

= \(m\left({ }_6^{11} \mathrm{C}\right)-m\left({ }_5^{11} \mathrm{~B}\right)-2 m_e\)

= (11.011434-11.009305-2 × 0.000548) u

= 0.001033 × 931.2 MeV

= 0.962 MeV

This energy is almost equal to the maximum energy released In the decay process

 

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