Updated By SaraNextGen

On April 21, 2024, 11:35 AM

Question 13.9:

Obtain the amount of  necessary to provide a radioactive source of 8.0 mCi strength. The half-life of  is 5.3 years.

Answer:

The strength of the radioactive source is given as:

Where,

N = Required number of atoms

Half-life of ,   = 5.3 years

= 5.3 × 365 × 24 × 60 × 60

= 1.67 × 10⁸ s

For decay constant λ, we have the rate of decay as:

Where, λ 

For :

Mass of 6.023 × 10²³ (Avogadro’s number) atoms = 60 g

∴Mass of  atoms 

Hence, the amount of   necessary for the purpose is 7.106 × 10⁻⁶ g.

Question 13.10:

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

Answer:

Half life of  ,  = 28 years

= 28 × 365 × 24 × 60 × 60

= 8.83 × 10⁸ s

Mass of the isotope, m = 15 mg

90 g of  atom contains 6.023 × 10²³ (Avogadro’s number) atoms.

Therefore, 15 mg of   contains:

Rate of disintegration, 

Where,

λ = Decay constant 

Hence, the disintegration rate of 15 mg of the given isotope is 7.878 × 10¹⁰ atoms/s.

Question 13.11:

Obtain approximately the ratio of the nuclear radii of the gold isotope   and the silver isotope .

Answer:

Nuclear radius of the gold isotope  = R_Au

Nuclear radius of the silver isotope  = R_Ag

Mass number of gold, A_Au = 197

Mass number of silver, A_Ag = 107

The ratio of the radii of the two nuclei is related with their mass numbers as:

Hence, the ratio of the nuclear radii of the gold and silver isotopes is about 1.23.

Question 13.12:

Find the Q-value and the kinetic energy of the emitted α-particle in the α-decay of (a)  and (b) .

Given

 = 226.02540 u,   = 222.01750 u,

= 220.01137 u,  = 216.00189 u.

Answer:

(a) Alpha particle decay of  emits a helium nucleus. As a result, its mass number reduces to (226 − 4) 222 and its atomic number reduces to (88 − 2) 86. This is shown in the following nuclear reaction.

Q-value of

emitted α-particle = (Sum of initial mass − Sum of final mass) c²

Where,

c = Speed of light

It is given that:

Q-value = [226.02540 − (222.01750 + 4.002603)] u c²

= 0.005297 u c²

But 1 u = 931.5 MeV/c²

∴Q = 0.005297 × 931.5 ≈ 4.94 MeV

Kinetic energy of the α-particle 

(b) Alpha particle decay of   is shown by the following nuclear reaction.

It is given that:

Mass of  = 220.01137 u

Mass of  = 216.00189 u

∴Q-value = 

≈ 641 MeV

Kinetic energy of the α-particle 

= 6.29 MeV

Question 13.13:

The radionuclide ¹¹C decays according to

The maximum energy of the emitted positron is 0.960 MeV.

Given the mass values:

calculate Q and compare it with the maximum energy of the positron emitted

Answer:

The given nuclear reaction is:

Atomic mass of  = 11.011434 u

Atomic mass of 

Maximum energy possessed by the emitted positron = 0.960 MeV

The change in the Q-value (ΔQ) of the nuclear masses of the   nucleus is given as:

Where,

me = Mass of an electron or positron = 0.000548 u

c = Speed of light

m’ = Respective nuclear masses

If atomic masses are used instead of nuclear masses, then we have to add 6 me in the case of and 5 me in the case of .

Hence, equation (1) reduces to:

∴ΔQ = [11.011434 − 11.009305 − 2 × 0.000548] c²

= (0.001033 c²) u

But 1 u = 931.5 Mev/c²

∴ΔQ = 0.001033 × 931.5 ≈ 0.962 MeV

The value of Q is almost comparable to the maximum energy of the emitted positron.

Question 13.14:

The nucleus  decays by emission. Write down the  decay equation and determine the maximum kinetic energy of the electrons emitted. Given that:

= 22.994466 u

= 22.989770 u.

Answer:

In   emission, the number of protons increases by 1, and one electron and an antineutrino are emitted from the parent nucleus.

 emission of the nucleus   is given as:

It is given that:

Atomic mass of  = 22.994466 u

Atomic mass of  = 22.989770 u

Mass of an electron, me = 0.000548 u

Q-value of the given reaction is given as:

There are 10 electrons in   and 11 electrons in . Hence, the mass of the electron is cancelled in the Q-value equation.

The daughter nucleus is too heavy as compared to  and  . Hence, it carries negligible energy. The kinetic energy of the antineutrino is nearly zero. Hence, the maximum kinetic energy of the emitted electrons is almost equal to the Q-value, i.e., 4.374 MeV.

Question 13.15:

The Q value of a nuclear reaction A + b → C + d is defined by

Q = [ mA+ mb− mC− md]c² 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) 

(ii) 

Atomic masses are given to be

Answer:

(i) The given nuclear reaction is:

It is given that:

Atomic mass 

Atomic mass 

Atomic mass 

According to the Question, the Q-value of the reaction can be written as:

The negativeQ-value of the reaction shows that the reaction is endothermic.

(ii) The given nuclear reaction is:

It is given that:

Atomic mass of 

Atomic mass of 

Atomic mass of 

The Q-value of this reaction is given as:

The positive Q-value of the reaction shows that the reaction is exothermic.

Question 13.16:

Suppose, we think of fission of a  nucleus into two equal fragments, . Is the fission energetically possible? Argue by working out Q of the process. Given   and .

Answer:

The fission of  can be given as:

It is given that:

Atomic mass of   = 55.93494 u

Atomic mass of 

The Q-value of this nuclear reaction is given as:

The Q-value of the fission is negative. Therefore, the fission is not possible energetically. For an energetically-possible fission reaction, the Q-value must be positive.

Question 13.17:

The fission properties of  are very similar to those of .

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  undergo fission?

Answer:

Average energy released per fission of , 

Amount of pure , m = 1 kg = 1000 g

N_A= Avogadro number = 6.023 × 10²³

Mass number of = 239 g

1 mole of  contains N_A atoms.

∴m g of  contains

∴Total energy released during the fission of 1 kg of is calculated as:

Hence,   is released if all the atoms in 1 kg of pure  undergo fission.