# Physics Class 12 NCERT Solutions: Chapter 13 Nuclei Part 8

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Q: 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 .)

Answer:

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

Radius of a deuteron nucleus

Charge on a deuteron nucleus Charge on an electron

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.

Q: 21. From the relation, where 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:

Where,

Let be the average mass of the nucleus.

Hence, mass of the nucleus

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

Q: 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. The nuclear reaction can be written as:

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

Nuclear mass of

Nuclear mass of

Atomic mass of

Atomic mass of

Mass of an electron

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, it does not necessarily mean that.

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 value must be positive for an energetically-allowed nuclear reaction.