Page No 464: - Chapter 13 Nuclei Exercise Solutions class 12 ncert solutions Physics - SaraNextGen [2024-2025]

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On April 24, 2024, 11:35 AM

Question 13.18:

A 1000 MW fission reactor consumes half of its fuel in 5.00 y. How much

did it contain initially? Assume that the reactor operates 80% of the time, that all the energy generated arises from the fission of  and that this nuclide is consumed only by the fission process.

Answer:

Half life of the fuel of the fission reactor,  years

= 5 × 365 × 24 × 60 × 60 s

We know that in the fission of 1 g of  nucleus, the energy released is equal to 200 MeV.

1 mole, i.e., 235 g of  contains 6.023 × 10²³ atoms.

∴1 g   contains

The total energy generated per gram of is calculated as:

The reactor operates only 80% of the time.

Hence, the amount of  consumed in 5 years by the 1000 MW fission reactor is calculated as:

∴Initial amount of  = 2 × 1538 = 3076 kg

Question 13.19:

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

Answer:

The given fusion reaction is:

Amount of deuterium, m = 2 kg

1 mole, i.e., 2 g of deuterium contains 6.023 × 10²³ atoms.

∴2.0 kg of deuterium contains 

It can be inferred from the given reaction that when two atoms of deuterium fuse, 3.27 MeV energy is released.

∴Total energy per nucleus released in the fusion reaction:

Power of the electric lamp, P = 100 W = 100 J/s

Hence, the energy consumed by the lamp per second = 100 J

The total time for which the electric lamp will glow is calculated as:

Question 13.20:

Calculate the height of the potential barrier for a head on collision of two deuterons. (Hint: The height of the potential barrier is given by the Coulomb repulsion between the two deuterons when they just touch each other. Assume that they can be taken as hard spheres of radius 2.0 fm.)

Answer:

When two deuterons collide head-on, the distance between their centres, d is given as:

Radius of 1^st deuteron + Radius of 2^nd deuteron

Radius of a deuteron nucleus = 2 fm = 2 × 10⁻¹⁵ m

∴d = 2 × 10⁻¹⁵ + 2 × 10⁻¹⁵ = 4 × 10⁻¹⁵ m

Charge on a deuteron nucleus = Charge on an electron = e = 1.6 × 10⁻¹⁹ C

Potential energy of the two-deuteron system:

Where,

 = Permittivity of free space

Hence, the height of the potential barrier of the two-deuteron system is

360 keV.

Question 13.21:

From the relation R = R₀A¹/³, where R₀ is a constant and A is the mass number of a nucleus, show that the nuclear matter density is nearly constant (i.e. independent of A).

Answer:

We have the expression for nuclear radius as:

R = R₀A¹/³

Where,

R₀ = Constant.

A = Mass number of the nucleus

Nuclear matter density, 

Let m be the average mass of the nucleus.

Hence, mass of the nucleus = mA

Hence, the nuclear matter density is independent of A. It is nearly constant.

Question 13.22:

For the   (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   emission is energetically allowed, electron capture is necessarily allowed but not vice−versa.

Answer:

Let the amount of energy released during the electron capture process be Q₁. The nuclear reaction can be written as:

Let the amount of energy released during the positron capture process be Q₂. The nuclear reaction can be written as:

= Nuclear mass of 

= Nuclear mass of 

= Atomic mass of 

= Atomic mass of 

me = Mass of an electron

c = Speed of light

Q-value of the electron capture reaction is given as:

Q-value of the positron capture reaction is given as:

It can be inferred that if Q₂ > 0, then Q₁ > 0; Also, if Q₁> 0, it does not necessarily mean that Q₂ > 0.

In other words, this means that if emission is energetically allowed, then the electron capture process is necessarily allowed, but not vice-versa. This is because the Q-value must be positive for an energetically-allowed nuclear reaction.