Question

Solution

Correct option is The distance of closest approach of an -particle (charge +2e) of initial kinetic energy K, scattering through 180o from a nucleus of atomic numberZ is given by Substituting the given value, we get  The radius of the gold nucleus is, therefore, less than (= 18 fermi). This is the upper limit of the radius of gold nucleus.

SIMILAR QUESTIONS

Q1

The ratio of ionization energy of hand is is

Q2

Potential energy of list electron is

Q3

The ratio of ionization energy of hydrogen atom in terms of Reedley constant is given

Q4

An -part ale of energy 5Mev is scattered through 180 by a fined uranium nucleus. The distance of the closed approach is of the order of:

Q5

An -particle with kinetic energy 10 MeV is heading towards a stationary point-nucleus of atomic number Z = 50. Calculate the distance of closest approach. Q6

An -particle with kinetic energy 10 MeV is heading towards a stationary point-nucleus of atomic number Z = 50. Calculate the distance of closest approach. Q7

What is the distance of closest approach when a 5.0 MeV proton approaches a gold nucleus (Z = 79)? Given,  .

Q8

An -particle after passing through a potential difference of falls on a silver foil. The atomic number of silver is 47. Calculate the kinetic energy of the -particle at the time of falling on the foil.

Q9

A beam of -particles of velocity is scattered by a gold (Z = 79) foil. Find out the distance closest approach of the -particle to the gold nucleus. The value of charge/mass for -particle is .

Q10

In a head-on collision between an -particle and gold (Z = 79) nucleus, the closest distance of approach is 41.3 fermi. Calculate the energy of the -particle. (1 fermi = 10–15 m)