Nuclear Physics

In decay, a neutron in the nucleus decays into a proton, an electron ( particle), and an antineutrino. The electron is emitted from the nucleus, which results in the conversion of the neutron into a proton. The process can be written as: Thus, during decay, a neutron in the nucleus decays emitting an electron, which is the characteristic feature of this type of radioactive decay.

The correct answer is (C) : A neutron in the nucleus decays emitting an electron.

In the case of alpha particles directed towards a thin film of gold, the scattering of the particles occurs due to the interference between the transmitted and reflected beams.
This scattering is a typical phenomenon observed in diffraction patterns, where the intensity of the scattered particles varies as a function of the angle.
When the incident beam passes through the thin film, interference results in maxima and minima for the transmitted and reflected parts of the beam.
, which corresponds to the transmitted part, will experience maximum intensity as it passes through the region of the film where constructive interference is most prominent.
, corresponding to a reflected part, will experience a minimum intensity because destructive interference occurs at this point.

The correct answer is (B) : A' will be maximum and in B' minimum.

• Repels strongly at small separations (r < 0.5 fm) → high positive U.
• Attracts at intermediate distances (0.5 fm < r < 2 fm) → negative U.
• Drops to zero for r > 2 fm.

Why D? Only option D uses correct units (U in MeV, r in fm) and matches this behavior.

Others fail because:

• B, C: Wrong units (r in meters, U in keV).
• A: Undefined graph shape (if drawn correctly, could also be valid).

The correct answer is (D) : .

Using the equation , we can solve for mass :

Given values:

• Power
• Speed of light
• Time

Substituting the given values into the equation:

Simplifying the equation:

Therefore, the mass is: 0.04 g

Correct Answer: D

Explanation:

The given equation represents a beta-minus (β^-) decay, where a neutron is converted into a proton, emitting an electron (e^-) and an antineutrino (ν_e).

The equation is as follows:

In β^- decay, a neutron in the nucleus of the atom is transformed into a proton. The atomic number (Z) increases by 1, while the mass number (A) remains the same because a neutron is converted to a proton without a change in the total nucleon count.

In the equation for bismuth-210 ( ), the atomic number is 83 and the mass number is 210. After the β^- decay, the resulting element will have an atomic number of 84 (one proton more than 83), but the mass number will remain the same (210). This corresponds to the element polonium (Po), which has atomic number 84 and mass number 210.

The number of neutrons in the resulting nucleus can be found by subtracting the atomic number from the mass number:

Neutrons = Mass number - Atomic number = 210 - 84 = 126

Thus, the number of neutrons in the nucleus X is 126, which corresponds to the correct answer D.

Correct Answer: A

Explanation:

In a nuclear reaction, when a nucleus emits an alpha particle, the total energy released in the reaction is given by the Q value of the reaction. The Q value represents the total energy released, which is shared as kinetic energy between the products of the reaction.

In this case, the nucleus has a mass number of 220 and is initially at rest. It emits an alpha particle, and the Q value of the reaction is given as 5.5 MeV.

The energy released is shared between the alpha particle and the daughter nucleus. Since the daughter nucleus is much heavier than the alpha particle, the kinetic energy of the alpha particle will be very close to the Q value. The difference arises due to the mass ratio between the alpha particle and the daughter nucleus, but the kinetic energy of the alpha particle is approximately equal to the Q value.

Thus, the kinetic energy of the alpha particle is 5.4 MeV, which corresponds to the correct answer A.

Step 1: Understanding Rutherford’s Alpha Scattering Experiment
In Rutherford’s alpha scattering experiment, the impact parameter ( ) is defined as the perpendicular distance between the direction of motion of the alpha particle and the center of the nucleus of the atom being targeted.

Step 2: Relation between Impact Parameter and Scattering Angle
The scattering angle ( ) is the angle through which the alpha particle deviates from its original path. As per Rutherford’s scattering theory:

• A small impact parameter (alpha particle approaching nucleus closely) leads to a large scattering angle (close to 180° for a head-on collision).
• A large impact parameter (alpha particle passing far away from nucleus) leads to a smaller scattering angle (particle barely deflected).

Thus, clearly as the impact parameter increases, the scattering angle decreases.