Question:

The distance of closest approach for an alpha nucleus of velocity $ v $ bombarding a stationary heavy nucleus target of charge $ Ze $ is directly proportional to

Updated On: Jun 2, 2024
  • $ v $
  • $ m $
  • $ \frac{1}{v^{2}} $
  • $ \frac{1}{Ze} $
Hide Solution
collegedunia
Verified By Collegedunia

The Correct Option is C

Solution and Explanation

At the distance of closest approach $(d)$,
Kinetic energy of $\alpha$-particle
= Potential energy of $\alpha$- particle and target nucleus
$\therefore \frac{1}{2}mv^{2}=\frac{1}{4\pi\varepsilon_{0}} \frac{\left(2e\right)\left(ze\right)}{d}$
or $d=\frac{1}{4\pi\varepsilon_{0}} \frac{4Ze^{2}}{mv^{2}}$
$\therefore d \propto\frac{1}{v^{2}}$
Was this answer helpful?
0
0

Top Questions on Nuclear physics

View More Questions

Concepts Used:

Nuclear Physics

Nuclear physics is the field of physics that studies atomic nuclei and their constituents and interactions, in addition to the study of other forms of nuclear matter. Nuclear physics should not be confused with atomic physics, which studies the atom as a whole, including its electrons

Radius of Nucleus

‘R’ represents the radius of the nucleus. R = RoA1/3

Where,

  • Ro is the proportionality constant
  • A is the mass number of the element

Total Number of Protons and Neutrons in a Nucleus

The mass number (A), also known as the nucleon number, is the total number of neutrons and protons in a nucleus.

A = Z + N

Where, N is the neutron number, A is the mass number, Z is the proton number

Mass Defect

Mass defect is the difference between the sum of masses of the nucleons (neutrons + protons) constituting a nucleus and the rest mass of the nucleus and is given as:

Δm = Zmp + (A - Z) mn - M

Where Z = atomic number, A = mass number, mp = mass of 1 proton, mn = mass of 1 neutron and M = mass of nucleus.