Question

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) 

Solution

Correct option is

5.5 MeV

The distance of closest approach of an -partcle of energy K is 

      .  

  

Putting the given known values, we get 

      

          

.       

  

           = 5.5 MeV.

SIMILAR QUESTIONS

Q1

Potential energy of list electron is

Q2

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

Q3

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:

Q4

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.

   

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

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

    

.

Q7

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. 

Q8

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 .

Q9

What is the upper limit of the radius of the gold nuclues (Z = 79), if an -particle of energy 12.5 MeV is deflected back by the nucleus through 180o.

Q10

Write down the expression for the radii of orbits of hydrogen atom. Calculate the radius of the smallest orbit.