List of top Physics Questions on Nuclear Physics asked in CBSE CLASS XII

The ‘distance of closest approach’ of an alpha-particle is \(d\) when it moves with a velocity \(v\) head-on towards the target nucleus. If the velocity of alpha particle is halved, the new ‘distance of closest approach’ will be –
The ‘distance of closest approach’ of an alpha-particle is \(d\) when it moves with a velocity \(v\) head-on towards the target nucleus. If the velocity of alpha particle is halved, the new ‘distance of closest approach’ will be –
Define ‘Mass defect’ and ‘Binding energy’ of a nucleus. Describe ‘Fission process’ on the basis of binding energy per nucleon.
A deuteron contains a proton and a neutron and has a mass of 2.013553 u. Calculate the mass defect for it in u and its energy equivalence in MeV.
Given:
\( m_p = 1.007277 \) u, \( m_n = 1.008665 \) u, \( 1 \) u = \( 931.5 \) MeV/\( c^2 \).
Which of the following figures correctly represents the shape of the curve of binding energy per nucleon as a function of mass number?
Assertion (A): An alpha particle is moving towards a gold nucleus. The impact parameter is maximum for the scattering angle of 180°. % Reason (R) Reason (R): The impact parameter in an alpha particle scattering experiment does not depend upon the atomic number of the target nucleus.
The potential energy between two nucleons inside a nucleus is minimum at a distance of about: