Gujarat Board GSEB Textbook Solutions Class 12 Physics Chapter 13 Nuclei Textbook Questions and Answers, Additional Important Questions, Notes Pdf.

## Gujarat Board Textbook Solutions Class 12 Physics Chapter 13 Nuclei

### GSEB Class 12 Physics Nuclei Text Book Questions and Answers

Question 1.

(a) Two stable isotopes of lithium ^{6}_{3}Li and ^{7}_{3}Li have respective abundances of 7.5% and 92.5%. These isotopes have masses 6.01512 u and 7.01600 u, respectively. Find the atomic mass of lithium.

(b) Boron has two stable isotopes ^{10}_{5}B and ^{11}_{5}B. Their respective masses are 10.01294 u and 11.00931 u and the atomic mass of boron is 10.811 u. Find the abundances of ^{10}_{5}B and ^{11}_{5}B.

Solution:

(a) The average atomic mass of lithium

(b) Suppose the natural boron contains x% of ^{10}_{5}B isotop and (100 – x)% of ^{11}_{5}B isotope.

Then atomic mass of natural boron = weighed average of the masses of two isotopes.

On solving, x = 19.9

∴Relative abundance of ^{10}_{5}B isotope = 19.9%

∴Relative abundance of ^{11}_{5}B isotope = 80.1%

Question 2.

The three stable isotopes of neon: ^{20}_{10}Ne, ^{21}_{10}Ne and ^{22}_{10}Ne have respective abundances of 90.51%, 0.27% and 9.22%. The atomic masses of the three isotopes are 19.99 u, 20.99 u and 21.99 u, respectively. Obtain the average atomic ms of neon.

Solution:

The average atomic mass of neon =

Question 3.

Obtain the binding energy (in MeV) of a nitrogen nucleus (^{14}_{7}N), given m(^{14}_{7}N) =14.00307 u.

Solution:

BE = [{Zm_{p} + (A – Z)m_{n}}-M]c^{2}

Binding energy of ^{14}_{7}N = [{7 x 1.007825 + 7 x 1.008665} – 14.00307]931(∴1u = \(\frac{\mathrm{MeV}}{\mathrm{c}^{2}}\) = 104.61 MeV

Question 4.

Obtain the binding energy of the nuclei Fe and 209131 in units of MeV from the following data.

m(^{56}_{26}Fe) = 55.934939 u, m(^{209}_{83}Bi) 208.980388 u

BE = [(Zm + (A – Z)m_{n}) – M]c^{2}

lu = 931 MeV

∴ BE of ^{56}_{26}Fe = [(26 x 1.007825 + 30 x 1.008665) – 55.934939] 931 MeV = 492 MeV

∴ BE/nucleon =\(\frac {492}{56} \) = 8.79 MeV

Similarly BE/nucleon of ^{209}_{83}Bi = 7.84 MeV

Question 5.

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 Cu atoms (of mass 62.92960 u).

Solution:

BE = {[Zm_{p} + (A – Z)m_{n} – M} 931 MeV

= [29 x 1.007825 + (63 – 29) x 1 .008665 – 62.92960] 931 = 551.38 MeV

Number of atoms ¡n 63 g of Cu = 6.023 x 10^{23}

∴ Number of atoms in 3 g of Cu = img = 0.2868095 x 10^{23}

∴ The energy required to separate all protons and neutrons,

E = BE x n = 551.38 x 0.2868095 x 10^{23} = 1.58 x 10^{25}MeV

Question 6.

Write nuclear reaction equations for

- α – decay of
^{226}_{88}Ra - α – decay of
^{242}_{94}pu - β
^{–}– decay of^{32}_{15}P - β
^{–}– decay of^{210}_{83}Bi - β
^{+}– decay of^{11}_{6}C - β
^{+}– decay of^{97}_{43}Tc - Electron capture of
^{120}_{54}Xe.

Answer:

^{226}_{88}Ra →^{222}_{86}Rn +^{4}_{2}He^{226}_{88}Pu →^{222}_{86}U +^{4}_{2}He^{226}_{88}P →^{222}_{86}S + e + \(\overline { \upsilon }\)^{226}_{88}Bi →^{222}_{86}Po + e^{–}+ \(\overline { \upsilon }\)^{11}_{6}C →^{11}_{5}B + e^{+}+ υ^{11}_{6}Tc →^{11}_{6}Mo + e^{+}+ υ^{120}_{54}Xe + e^{–}→^{120}_{54}I + υ

Question 7.

A radioactive isotope has a half-life of T years. How long will it take the activity to reduce to (a) 3.125%, (b) 1% of its original value?

Solution:

(a) \(\frac{\mathrm{A}}{\mathrm{A}_{0}}\) = 3 .125% = \(\frac{3.125}{100}\)

\(\frac{\mathrm{A}}{\mathrm{A}_{0}}\) = \(\frac{100}{3.125}\) = 32

\(\left(\frac{\mathrm{A}_{0}}{\mathrm{~A}}\right)\) = 2^{n} i.e., 32 = 2^{n}

n log 2 = 1og 32, n = \(\frac{\log 32}{\log 2}\) = 5

\(\frac{t}{T}\) = n = 5, t = 5T

(b) \(\frac{\mathrm{A}}{\mathrm{A}_{0}}\) = \(\frac{1}{100}\),

\(\frac{A_{0}}{A}\) = 100 = 2^{n}

n log 2 = log 100

\(\frac{t}{T}\) = n = \(\left(\frac{\log 100}{\log 2}\right)\),

∴ t = 6.65 T

Question 8.

The normal activity of living carbon-containing matter is found to be about 15 decays per minute for every gram of carbon. This activity arises from the small proportion of radioactive ^{14}_{6}C present with the stable carbon isotope ^{12}_{6}C. When the organism is dead, its interaction with the atmosphere (which maintains the above equilibrium activity) ceases and its activity begins to drop. From the known half-life (5730 years) of ^{14}_{6}C, and the measured activity, the age of the specimen can be approximately estimated. This is the principle of ^{14}_{6}C dating used in archaeology. Suppose a specimen from Mohen jodaro gives an activity of 9 decays per minute per gram of carbon. Estimate the approximate age of the Indus-Valley civilisation.

Solution:

Normal activity, R_{0} = 15 decays per min.

Present activity. R = 9 decays per min.

\(\mathrm{T}_{1 / 2}\) = 5730 days

N = N_{0}e^{-λt}

e^{-λt} = \(\frac{\mathrm{R}}{\mathrm{R}_{0}}\) = \(\frac{9}{15}\)

λt = log_{e}\(\frac{15}{9}\) = 2.303 l0g_{10}\(\left(\frac{5}{9}\right)\) = 0.511

t = \(\frac{0.511}{\lambda}\) = \(\frac{0.511}{0.693}\) x \(\mathrm{T}_{1 / 2}\)

= \(\frac{0.511}{0.693}\) x 5730 = 4225 years

Question 9.

Obtain the amount of ^{60}_{27}Co necessary to provide a radioactive source of 8.0 mCi strength. The half-life of ^{60}_{27}Co is 5.3 years.

Solution:

Activity, A = 8 mCi = 8 x 10^{-3} Ci = 8 x 10^{-3} x 3.7 x 10^{10} = 2.96 x 10^{8}s^{-1}

\(\mathrm{T}_{1 / 2}\) = 5.3 years

5.3 x 365 x 24 x 60 x 60 = 1.67 x 10^{8} s

λ = \(\frac{0.693}{\mathrm{~T}_{1 / 3}}\) = \(\frac{0.693}{1.67 \times 10^{8}}\) = 4.15 x 10^{-9}s^{-1}

A = λ N

N = \(\frac{A}{λ}\) = \(\frac{2.96 \times 10^{8}}{4.15 \times 10^{-9}}\) = 7.13 x 10^{16}

Mass of 60 g of Co atoms = \(\frac{60 \times 7.13 \times 10^{16}}{6.02 \times 10^{23}}\) = 7.13 x 10^{16}g

Question 10.

The half-life of ^{90}_{38}Sr is 28 years. What is the disintegration rate of 15 mg of this isotope?

\(\mathrm{T}_{1 / 2}\) = 28 years = 28 x 365 x 24 x 60 x 60 s = 8.83 x 10^{8} s

Solution:

Question 11.

Obtain approximately the ratio of the nuclear radii of the gold isotope ^{197}_{79}Au and the silver isotope ^{107}_{47}Ag.

Solution:

R = R_{0}\(\mathrm{R}_{0} \mathrm{~A}^{1 / 3}\)

For ^{197}_{79}Au, R_{Au} = R_{0}\((197)^{1 / 3}\)

For ^{107}_{47}Ag, R _{Ag} = R_{0}\((107)^{1 / 3}\)

\(\frac{\mathrm{R}_{\mathrm{Au}}}{\mathrm{R}_{\mathrm{Ag}}}\) = \(\left(\frac{197}{107}\right)^{1 / 3}\) = 1.23

Question 12.

Find the Q – value and the kinetic energy of the emitted α – particle in the α – decay of

(a) ^{226}_{88}Ra (b) ^{220}_{86}Rn

Given m (^{226}_{88}Ra) = 226.02540 u’

m(^{222}_{86}Rn) = 222.01750 u,

m(^{220}_{86}Rn) = 220.01137 u,

m (^{216}_{84}Po) = 216.00189 u.

Solution:

Q = E_{α}\(\left(\frac{A}{A-4}\right)\) where E_{α} is the KE of α particle.

A – mass number of parent

(a) ^{226}_{88}Ra → ^{222}_{86}Ra + α + Q

226.02540 – (222.01750 + 4.002603) = 0.005 5297

∴Q = 0.005297 x 931.5 = 4.934 MeV

∴E_{α} = 4.934 x \(\frac{A-4}{A}\) = 4.934 \(\frac{222}{226}\) = 4.85 MeV

(b) Solve following similar procedure

Question 13.

The radionuclide “C decays according to

^{11}_{6}C →^{11}_{5}B + e^{+} + υ, \(\mathrm{T}_{1 / 2}\) = 20.3 min

The maximum energy of the emitted positron is 0.960 MeV.

Given the mass values: m (^{11}_{6}C) = 11.011434 u and m (^{11}_{6}B) = 11.009305 u. Calculate Q and compare it with the maximum energy of the positron emitted.

Solution:

The disintegration energy.

Q = (m_{C} – m_{B} – 2m_{c})c^{2} = (m_{C} – m_{B} – 2m_{c}) \(\frac{931.5 \mathrm{c}^{2}}{\mathrm{c}^{2}}\) MeV

2m_{e} = 2 x 0.0005486 = 0.0010972

Since the maximum energy of positron emitted is 0.960 MeV, the neutrino carries only negligible enegry

Question 14.

The nucleus ^{23}_{10}Ne decays by β^{–} emission. Write down the β-decay equation and determine the maximum kinetic energy of the electrons emitted. Given that:

m (^{23}_{10}Ne) = 22.994466 u

m (^{23}_{11}Na) = 22.089770 u

Solution:

The equation is ^{23}_{10}Ne →^{23}_{11}Na + β + \(\overline { \upsilon }\) + Q

Q = [m_{n}(^{23}_{10}Ne) + m_{n}^{23}_{11}Na + 10m_{e}]c^{2}

(Neglect the rest mass of \(\overline { \upsilon }\))

= [{m_{n} ^{23}_{10}Ne + 10m_{e}} – {m_{n}(^{23}_{11}Na) + 11m_{e}}]c^{2}

= [m ^{23}_{10}Ne – m ^{23}_{11}Na]c^{2}

= [22 .994466 – 22.989770]931

= 4.374 MeV

Question 15.

The Q value of nuclear reaction A+ b → C + d is defined by Q = [m_{A} + m_{b} – m_{C} – m_{d}]c^{2} where the masses refer to the respective nuclei. Determine from the given data the Q-value of the following reactions and state whether the reactions are exothermic or endothermic.

I. ^{1}_{1}H + ^{3}_{1}H→ ^{2}_{1}H + ^{2}_{1}H

II. ^{12}_{6}C +^{12}_{6}C – ^{20}_{10}Ne + ^{4}_{2}He

Atomic masses are given to be

m (^{2}_{1}H) = 2.014102 u

m (^{3}_{1}H) = 3.016049 u

m (^{12}_{6}C) = 12.000000 u

m (^{20}_{10}Ne) = 19.992439 u

Solution:

I. ^{1}_{1}H + ^{3}_{1}H→ ^{2}_{1}H + ^{2}_{1}H

Q = [{m(^{2}_{1}H) + m(^{3}_{1}H)} – {m(^{2}_{1}H) + m( ^{2}_{1}H)}]c^{2}

=(4.02387 – 4.028204)931

= – 4.033 MeV (Negative sign shows that the reaction is endothermic)

II. Q [2m(^{12}_{6}C) – {m(^{20}_{10}Ne) + m(^{4}_{2}He)}]c^{2}

= 2 x 12 – (19.992439 + 4.002603)931

4.618 MeV (exothermic reaction)

Question 16.

Suppose, we think of fission of a ^{56}_{26}Fe nucleus into two equal fragments., ^{28}_{13}Al. is fission energetically possible? Argue by working out Q of the process. Given m ( ^{56}_{26}Fe = 55.93494 u and m (^{28}_{13}Al) = 27.98191 u.

Solution:

Q = [m ^{56}_{26}Fe – 2m(^{28}_{13}Al)]c^{2}

= [55.93494 – 2 x 27.98191]931

= 26.88 MeV (not possible)

Question 17.

The fission properties of ^{239}_{92}Pu are very similar to those of ^{239}_{92}U The average enegry released per fission is 180 MeV. How much energy, in MeV, is released if all atoms in 1 kg of pure ^{239}_{92}Pu undergo fission?

Solution:

Number of atoms in 239 g of 2 ^{239}_{92}Pu is Avagadro number.

Number of atoms in 1 g = \(\frac {N}{239} \) = \(\frac{6.023 \times 10^{23}}{239}\)

Number of atoms in 1 g = \(\frac{6.023 \times 10^{23}}{239}\) x 10^{3} = 2.52 x 10^{24} atoms of ^{239}_{92}Pu

Energy released during 1 kg of ^{239}_{94}pu = Energy released by 2.52 x 10^{24} atoms of ^{239}_{94}Pu

= Energy released per fission x Total number of atoms in 1 kg = 180 x 2.52 x 10^{24} = 4.536 x 10^{26} MeV

Question 18.

A 1000 MW fission reactor consumes half of its fuel in 5.00 yrs. how much ^{235}_{92}U did it contain initially? Assume that the reactor operates 80% of the time, that all the energy generated arises from the fission of ^{235}_{92}U and that this nuclei is consumed only by the fission process.

Solution:

The total power generated in 5 yrs. is

Let m kg be originally present. Then number of atoms undergone fission is that in \(\frac {m}{2} \)kg.

235 x 10^{-3} kg contains 6.023 x 10^{23} atoms.

Question 19.

How long can an electric lamp of 100 W be kept glowing by fusion of 2.0 kg of deuterium? Take the fusion reaction as

^{2}_{1}H + ^{2}_{1}H → ^{3}_{2}He + n + 3.27 MeV.

Solution:

Mass of deuterium = 2.0141 x 2 = 4.0282 u = 4.0282 x 1.66 x 10^{-27} kg

No. of fusion in kg of deuterium img = 2.991 x 10^{-26}kg

Energy released per fusion 3.2 MeV = 3.2 x 10^{6} x 1.6 x 10^{-19} x 2.991 x 10^{26}J

∴ Energy released from 2 kg of ^{2}_{1}H = E = 3.2 106 x 1.6 x x 2.991 x j

= 1.5314 x 10^{-14} J

E = P x t

Question 20.

For the β^{+} (positron) emission from a nucleus, there is another competing process known as electron capture (electron from an inner orbit, say, the K-shell, is captured by the nucleus and a neutrino is emitted).

Show that if β^{+} emission is energetically allowed, electron capture is necessarily allowed but not vice – versa.

Solution:

Consider the competing processes:

Question 21.

In a periodic table, the average atomic mass of magnesium is given as 24.312 u. The average value is based on their relative natural abundance on earth. The three isotopes and their masses are ^{24}_{12}Mg (23.98504 u), ^{25}_{12}Mg(24.98584 u) and ^{26}_{12}Mg(25.98259 u).

The natural abundance of ^{24}_{12}Mg is 78.99% by mass. Calculate the abundances of other two isotopes.

(^{24}_{12}Mg) = 24.312 u

Mass of ^{24}_{12}Mg = 23.985044

^{25}_{12}Mg = 24.98584 u

^{26}_{12}Mg = 25.98259 u

Let us assume that there are n_{1} atoms of 1^{25} isotope, n_{2} atoms of 2^{nd }etc. Total (n_{1} + n_{2} + n_{3}) hence (n_{1} + n_{2} + n_{3}) 24.312 gives the total mass.

Relative abundance of first = \(\frac{n_{1}}{\left(n_{1}+n_{2}+n_{3}\right)}\)

Relative abundance of 2^{nd} = \(\frac{n_{2}}{\left(n_{1}+n_{2}+n_{3}\right)}\)

Relative abundance of 3^{rd} = \(\frac{n_{3}}{\left(n_{1}+n_{2}+n_{3}\right)}\)

∴ n_{1} x 23.98504 + n_{2} x 24.98584 + n_{3} x 25.98259 =(n_{1} + n_{2} + n_{3}) x 24.3124

Sothat \(\frac{n_{1}}{\left(n_{1}+n_{2}+n_{3}\right)}\) x 23.98504 + \(\frac{n_{2}}{\left(n_{1}+n_{2}+n_{3}\right)}\) x 24.98584 + \(\frac{n_{3}}{\left(n_{1}+n_{2}+n_{3}\right)}\) x 25.98259 = 24.3126

\(\frac{n_{1}}{\left(n_{1}+n_{2}+n_{3}\right)}\)= 0.7899.

\(\frac{n_{2}}{\left(n_{1}+n_{2}+n_{3}\right)}\) = x,

\(\frac{n_{3}}{\left(n_{1}+n_{2}+n_{3}\right)}\) = 0.2101 – x

∴ 0.7899 x 23.98504 + x x 24.98584 + 0.2101 x 25.98259 – x x 25.98259 = 24.3126

18.9457515 + 5.458942153 – 24.3126 = (25.98259 – 24.98584)x

x = \(\frac{0.092093653}{0.99675}\) = 0.09239

i.e., 9.24%

Abundance of last isotope = 0.1177

. i.e., 11.77

Question 22.

The neutron separation energy is defined as the energy required to remove a neutron from the nucleus. Obtain the neutron separation energies of the nuclei ^{41}_{20}Ca and ^{27}_{13}Al from the following data:

m(^{40}_{20}Ca) = 39.962591 u

m(^{41}_{20}Ca) = 40.962278 u

m(^{26}_{13}Al) = 25.986895 u

m(^{27}_{13}A1) = 26.981541 u

Solution:

Neutron separation energy S_{n} of a nucleus ^{A}_{z}X is given by

Question 23.

A source contains two phosphorus radio nuclides ^{32}_{15}p (\(\mathrm{T}_{1 / 2}\) = 14.3 d) and ^{33}_{15}p (\(\mathrm{T}_{1 / 2}\) = 25.3 d). Initially, 10% of the decays come from ^{33}_{15}P . How long one must wait until 90% do so?

Solution:

Question 24.

Under certain circumstances, a nucleus can decay by emitting a particle more massive than an a -particle. Consider the following decay processes:

Calculate the Q-values for these decays and determine that both are energetically allowed.

Solution:

Question 25.

Consider the fission of ^{238}_{92}U by fast neutrons. In one fission event, no neutrons are emitted and the final end products, after the beta decay of the primary fragments, are ^{140}_{58}Ce and ^{99}_{44}Ru. Calculate Q for this fission process. The relevant atomic and particle masses are

m(^{238}_{92}U) = 238.05079 u

m(^{140}_{58}Ce) = 139.90543 u

m(^{94}_{44}Ru) = 98.90594 u

Q = [m(^{238}_{92}U) + m_{n} – m(^{140}_{58}Ce) – m(^{99}_{44}Ru)]c^{2}

= [238.05079 + 1.008665 – 139.90543 – 98.90594]931 = 231.1 MeV

Question 26.

Consider the D – T reaction (deuterium – tritium fusion)

^{2}_{1}H + ^{3}_{1}H → ^{4}_{2}He + n

(a) Calculate the energy released in MeV in this reaction from the data:

m(^{2}_{1}H) = 2.014102 u

m(^{3}_{1}H) = 3.016049 u

(b) Consider the radius of both deuterium and tritium to be approximately 2.0 fm. What is the kinetic energy needed to overcome the coulomb repulsion between the two nuclei? To what temperature must the gas be heated to initiate the

reaction?

(Hint: Kinetic energy required for one fusion event = average thermal kinetic energy available with the interacting particles = 2\(\left(\frac{3 \mathrm{kT}}{2}\right)\) ; k = Boltzman’s constant, T = absolute temperature.)

Solution:

(a) Q = m(^{2}_{1}H) + m(^{3}_{1}H) – m(^{4}_{1}He) + m_{n}]c^{2} = (2.014102 + 3.016049 – 4.002603 – 1.008665)931 = 17.59MeV

(b) Repulsive potential energy. E_{R}

Classically, a KE atleast equal to the above amount is required to overcome coulomb repulsion

KE = E_{k} = 7.68 x 10 ^{-14}

KE = \(\frac {3}{2} \)kT = E_{k}

Question 27.

Obtain the maximum kinetic energy of f3 particles, and the radiation frequencies of y decays in the decay scheme shown in figure. You are given that

m(^{198}Au) = 197.968233 u

m(^{198}Hg) = 197.966760 u

Solution:

Question 28.

Calculate and compare the energy released by (a) fusion of 1.0 kg of hydrogen deep within Sun and (b) the fission of 1.0 kg of ^{235}U in a fission reactor.

Solution:

(a) 1 g of ^{1}_{1}H contains Avogadro’s number

∴1kg of ^{1}_{1}H contains 6.023 x 10^{23} x 10^{3}

4 ^{1}_{1}H is required for a reaction, hence

Question 29.

Suppose india had a target of producing by 2020 AD, 200,000 MW of electric power, ten percent of which was to be obtained from nuclear power plants. Suppose we are given that, on an average, the efficiency of utilization (i.e. conversion to electric energy) of thermal energy produced in a reactor was 25%. How much amount of fissionable uranium would our country need per year by 2020? Take the heat energy per fission of ^{235}U to be about 200MeV.

Solution:

Power required from reactor ~2 x 10^{10} W

For 1 year it is 2 x 10^{10}) x 365.25 x 86400

Let m kg of uranium is required.

From 1 kg the energy released is

### GSEB Class 12 Physics Nuclei Additional Important Questions and Answers

Question 1.

How can you find the age of fossils?

Answer:

Using carbon dating

Question 2.

What is the significance of BE inside the nucleus?

Answer:

It holds the nucleons together inside the nucleus.

Question 3.

What happens lo the stability of nucleus, when BE increases?

Answer:

Nucleus becomes more stable.

Question 4.

How does nuclear fusion occur. even though coulombic repulsion exists?

Answer:

Nuclear force is stronger than the Coulombic force.

Question 5.

Nuclear fusion is called thermonuclear reaction. Why?

Answer:

Since the reaction takes place at a very high temperature of the order of million degree celsius. it is called thermonuclear reaction.

Question 6.

Classify the following statements into true or false.

- Radioactivity j.c unaffected by pressure and temperature.
- Radioactivity is affected by large electric and magnetic fields.
- The rate of disintegration of a radioactive substance at a particular rime is directly proportional to the number of atoms present initially.
- Nuclear force is charge independent but it depends on the relative orientation of the spins.

Answer:

- True
- False
- False
- True

Question 7.

Match the following

α -Partide | Electromagnetic waves | Speed is greater than c |

β – Particle | Fast moving neutrons | Speed is equal to c |

γ – rays | Nuclei of helium atom | Speed is aqual to 0.66c |

Fast moving electrons | Speed is about \(\frac{1}{10}\) velocity of light |

Answer:

α -Partide | Nuclei of helium atom | Speed is about \(\frac{1}{10}\) velocity of light |

β – Particle | Fast moving electrons | Speed is aqualt 0.66c |

γ – rays | Electromagnetic waves | Speed is equal to c |

Question 8.

(a) Is there any well defined boundary for a nucleus?

(b) What is the order of the size of a nucleus?

(c) How will you conclude that an atom of has a lot of empty space?

Answer:

(a) No

(b) Fermi

(c) Diameter of atom (including position of electrons) is of the order (≈) of 10^{-10}m and that of

nucleus is 10^{-15}m. The ratio is 10^{5} So much empty space ¡s present.

Question 9.

(a) What are ‘P ‘ and ‘Q’?

(b) What are the specialties of P?

Answer:

(a) P-neutron (^{1}_{0}n)

Q – energy

(b) i. A free neutron is unstable.

ii. Except hydrogen, it is present in the nucleus of all elements.

iii. Being a changeless particle, it is deflected by electric and magnetic fields.

iv. Energized neutron has high penetrating power.

Question 10.

The radius of a nucleus is R.

(a) How is it related to atomic mass (A)?

(b) Give the relation.

(c) What is the value of the constant of proportionality?

Answer:

(a) \(\mathrm{R} \propto \mathrm{A}^{\frac{1}{2}}\)

(b) \(\mathrm{R}=\mathrm{R}_{0} \mathrm{~A}^{\frac{1}{3}}\)

(c) R_{0} = (1.2 ±0.2) x 10^{-15} m

Question 11.

You have ^{11}_{6}C,^{12}_{6}C and ^{13}_{6}C.

(a) What are these nuclides?

(b) How do they differ?

Answer:

(a) They are isotopes

(b) In their mass numbers with difference in number of the neutrons.

Question 12.

We have ^{39}_{19}K and ^{37}_{17}Cl.

(a) What are these nuclides?

(b) How do they differ?

Answer:

(a) They are isotones

(b) May have the same number of neutron but differ in their atomic number and atomic mass.

Question 13.

There are two nuclides ^{22}_{10}Ne and ^{22}_{11}Na.

(a) What are these nuclides?

(b) How do they differ?

Answer:

(a) They are isobars

(b) Same number of nucleons but different number of protons

Question 14.

What are Isomers? Give examples.

Answer:

Isomers are nuclides with same number of protons and neutrons, but having different energy states and spins. They have radioactive property.

eg: ^{99}Tc & ^{99m}Tc, m – stands for meta stable state & Tc – Technetium

Question 15.

(a) What does the graph represent?

(b) What is the importance of this graph?

(c) What does the peak in the graph represent?

Answer:

(a) Binding energy/nucleon with atomic mass.

(b) It shows the stability of atomic nucleus

(c) The maximum stable condition of an element.

Question 16.

In certain isobars the number of protons of one isobar is equal to the number of protons in another.

(a) What is the name given to such isobars?

(b) Give one example.

Answer:

(a) Isomers

(b) ^{99}Tc & ^{99m}Tc

Question 17.

Z. A and M represent the atomic number, mass number and rest mass of a nucleus.

(a) Show that ‘M is always less than the mass of the constituent part ides.

(b) What is this mass difference called?

(c) Give the relation.

Answer:

(a) Protons and neutrons come together within a very small space of the order of 10^{-14}m to form a nucleus. The energy required to do so is provided by nucleus at the expense of their masses.

Hence the reason.

e.g., For _{5}B^{10}, the mass, M = 10.012944

_{5}B^{10} consists of 5 protons and 5 neutrons.

Hence total mass m_{p} + m_{n}

= 5 x Proton mass + 5 x Neutron mass

= 5 x 1.007825 + 5 x 1.008665

= 5.039125 + 5.043325 = 10.08245 u

i.e., M <(m_{p} + m_{N}), hence the reason.

(b) This mass difference is called mass defect (∆m)

(c) ∆m = Zm_{p} + (A – Z)m_{p} – M

Question 18.

(a) What are \({ }_{-1}^{0} \mathrm{e}\) and \(\bar { \upsilon }\) ?

(b) What ¡s the name given to this type of emission?

Answer:

(a) β-particle and anti-neutrino

(b) β-decay

Question 19.

A nuclear even is shown in the diagram.

(a) Name the nuclear phenomenon.

(b) What happens to the residual nucleus? Explain.

(c) Compare ¡he binding energy per nucleon of the daughter and the parent.

(d) At what time the hυ is emitted? Explain.

Answer:

(a) Radioactivity

(b) It rebounds (or recoils): due to momentum conservation principle.

(c) The BE per nucleon increases by the activity.

(d) There is no specific time. It is spontaneous. γ -emission can be independent or followed by α – decay

Question 20.

The radioactive decay of an element is shown in the graph.

(a) What is the sort of functional decay shown here?

(b) What is the change in activity in equal intervals of time?

(c) Find the half-flfe.

(d) Can you find the decay constant? if so, what is its value?

Answer:

(a) Hint: Apply λ = \(\frac{0.931}{\mathrm{~T}_{\frac{1}{2}}}\)

(b) Exponential decay

(c) y decreases by the same ratio

(d) Approximately 70 s.

Question 21.

Mention the different units of radioactivity

Answer:

Becquerel (Bq), Curie (Ci), Rutherford (Rf)

1 Bq = 1 disintegration/sec

1 Ci = 3.7 x 10^{10} disintegration/sec

1 Rf = 10^{6} disintegration/sec

Question 22.

Why are neutrons very effective as bombarding particles?

Answer:

As neutrons do not have any charge, they are not repelled by nucleus. Hence neutrons are more effective bombardment particles compared to charged particles.

Question 23.

(a) What are the different types of emission taking place here?

(b) Name the product after each emission.

(c) Juslify your answer

Answer:

(a) α – emission, β – emission and γ – emission

(c) When α – emission takes place, atomic number (Z) decreases by 2 and mass number (A) decreases by 4 and we get B. When a ^{176}_{70}B-emission takes place, Z increases by I and A remains the same. Thus we get ^{176}_{71}C. Again an β-emission takes place and we get ^{172}_{69}D.

Lastly γ – emission takes place and there is no change for Z and A. Thus we get ^{172}_{69}E.

Question 24.

From the above lubie we can see that the ratio of their atomic masses is 2. But it ¡s not exactly 2, Why is it so? Explain.

Answer:

Because the mass of a nucleus is slightly less than the mass of the constituent nucleons. (Actual atomic mass is different from mass number)

Question 25.

(a) What is a positron:? b. What is its charge?

Answer:

(a) A positron is the antiparticle of electron

(b) Same as that of electron but +ve.

Question 26.

What is

i. neutrino

ii. antineutrino?

Explain with examples.

Answer:

i. -β (i.e.. electron) emission is along with the emission of an antineutrino.

ii. +β (positron) emission is along with neutrino.

Neutrino and antineutrino have no charge and no mass.

Question 27.

α – particles have a high ionizing power. Justify our answer.

Answer:

α – particles have a charge of + 3.2 x 10^{-19}C and resembles helium nuclues.

Question 28.

(a) Are the equations of nuclear reactions such as ‘balanced’ in the sense a chemical equation If not, in what sense are they balanced on both sides?

(b) If the number of protons and number of neutrons are conserved in a nuclear reaction, in what way the mass’ is converted Into energy (or vice versa) in a nuclear reaction?

(c) A general impress ion exists that mass energy inter-conversion takes place only in nuclear reaction and never in chemical reaction. Strictly speaking, this is incorrect. Explain.

Answer:

(a) The total number of atoms remains the same in a chemical reaction. The number of atoms of each element is not necessarily conserved in a nuclear reaction.

(b) The binding energy (total) of the elements before the nuclear reaction need not be equal to the total binding energy after the reaction.

(c) There is mass defect in chemical reaction but it is almost a million times smaller compared with nuclear reaction.

Question 29.

Complete the nuclear reaction

Answer:

Question 30.

(a) Which is more stable, ^{7}_{3}Li or ^{4}_{3}Li?

(b) Give reason for your answer

Answer:

(a) ^{7}_{3}Li, is-stable

(b) It contains large number of neutrons than the other.

Question 31.

(a) What is the working principle behind

i. Atom bomb

ii. Hydrogen bomb

(b) What is meant by chain reaction?

(c) What happens if it is not controlled?

(d) Is there any device to control it?

Answer:

(a) i. Atom bomb – Nuclear fission

ii. Hydrogen bomb – Nuclear fusion

(b) Chain reaction is a series of fission process.

(c) A huge amount of energy is released which will lead the world to a disaster.

(d) Yes

Question 32.

(a) Are heavy nuclei stable or unstable?

(b) Give reason.

Answer:

(a) Unstable

(b) Binding energy per nucleon is small for heavy nuclei.

Question 33.

You are given two nucleoside ^{7}_{3}X and ^{4}_{3}Y.

(a) Are they the isotopes of the same element?

(b) Which one of the two is likely lo he more stable?

(c) Give reason.

Ans.

(a) Both arc isotopes of the same element since they have the same atomic number.

(b) ^{7}_{3}x

(c) Nuclides with greater number of neutrons is stable. So here ^{4}_{3} X is more stable than ^{7}_{3}Y.

Question 34.

What is the ratio of nuclear densities of two nucleus having mass numbers in the ratio 1:2?

Answer:

The nuclear density is independent of mass number. So the ratio is one.

Question 35.

(a) Give the expression for the radius of a nucleus.

(b) What is the rutio of the nuclear radii of ^{197}_{79}Au and ^{107}_{47}Ag?

(c) What is the ratio of their nuclear densities?

Answer:

(a) R = \(\mathrm{R}_{0} \mathrm{~A}^{\frac{1}{3}}\)

(b) \(\mathrm{R}_{0} \mathrm{~A}^{\frac{1}{3}}\)

(c) Registration

Question 36.

(a) is free neutron a stable particle?

(b) Give reason.

Answer:

(a) No

(b) A free neutron spontaneously decays into a proton, electron and an antineutrino.

Question 37.

(a) Give two radioactive elements which are not formed in observable quantities in nature.

(b) Give reason for this.

Answer:

(a) Plutonium, Tritium

(b) They have very small half-life period. Hence the reason.

Question 38.

What is positive β – emission?

Answer:

When a neutron splits into a proton, neutrino and positron, the emission is called positive β emission.

Question 39.

The mechanism of β emission can be stated as a neutron decay. Explain.

Answer:

A neutron decays into a proton (^{1}_{1}H) and an electron (^{0}_{-1}e). β-emission is the emission of electrons from the nucleus.

Actually there is an emission of a third particle called anti-neutrino (\(\bar { \upsilon }\)) in the β-emission process, so that both momentum and energy to be conserved.

Question 40.

Which raw material is used in a fast breeder reactor?

Answer:

Thorium

Question 41.

Hermes water is used as a moderator in a nuclear reactor Give reason

Answer:

Heavy water does not have a tendency to absorb neutrons. It slows down the fast moving neutrons.

Question 42.

What are

(a) slow neutrons?

(b) thermal neutrons?

(c) fast neutrons?

Answer:

(a) Slow neutrons → energy less than 1 eV

(b) Thermal neutrons → energy range 1 – 1.2 eV

(c) Fast neutrons → energy greater than 1.2 eV

Question 43.

(a) What factors make a fusion reaction difficult ¡o achieve?

(b) Why does a fusion reactor produce less radioactive waste than a fission reactor?

Answer:

(a) Nuclear fission occurs at a high temperature of 10^{5} – 10^{6}K. It is difficult to attain such a temperature.

(b) Whether it is fission or fusion, the product obtained is radio active. For a certain output energy the number of fusion reaction should he correspondingly large.

Question 44.

What is meant by a self sustained nuclear reaction?

Answer:

If the neutrons released in a nuclear fission can be used to promote further fission, then the reaction is a self sustained nuclear reaction.