Home
Class 12
PHYSICS
In the original experiment, Geiger and M...

In the original experiment, Geiger and Marsden calculated the distance of closest approach to the gold nucleus (Z=79)- of a 7.7MeV `alpha` particle before it comes momentarily to rest and reverses its direction. What is its value?

Text Solution

Verified by Experts

The key idea here is that throughout the scattering process, the total mechanical energy of the system consisting of an `alpha`-particle and a gold nucleus is conserved. The system’s initial mechanical energy is `E_(i)`, before the particle and nucleus interact, and it is equal to its mechanical energy `E_(f)` when the `alpha`-particle momentarily stops. The initial energy `E_(i)` is just the kinetic energy K of the incoming - particle. The final energy `E_(f)` is just the electric potential energy U of the system. The potential energy U can be calculated from Eq. (12.1).
Let d be the centre-to-centre distance between the `alpha`-particle and the gold nucleus when the `alpha`-particle is at its stopping point. Then we can write the conservation of energy `E_(i) = E_(f)` as
`K=(1)/(4piepsilon_(0))((2e)(Ze))/(d)=(2Ze^(2))/(4piepsilon_(0)d)`
Thus the distance of closest approach d is given by
`d=(2Ze^(2))/(4piepsilon_(0)K)`
The maximum kinetic energy found in `alpha`-particles of natural origin is 7.7 MeV or `1.2 xx 10^(12)`J. Since `1//4piepsilon_(0)=9.0xx10^(9)Nm^(2)//C^(2)`. Therefore with e = `1.6 xx 10^(-19)C`, we have,
`d=((2)(9.0xx10^(9)Nm^(2)//C^(2))(1.6xx10^(-19)C)^(2)Z)/(1.2 xx 10^(-12)J)`
`=3.84 xx 10^(-16) Zm`
The atomic number of foil material gold is Z = 79, so that d (Au) = `3.0 xx 10^(-14)m` = 30 fm. (1 fm (i.e. fermi) = `10^(-15)` m.)
The radius of gold nucleus is, therefore, less than `3.0 × 10^(–14)` m. This is not in very good agreement with the observed result as the actual radius of gold nucleus is 6 fm. The cause of discrepancy is that the distance of closest approach is considerably larger than the sum of the radii of the gold nucleus and the `alpha`-particle. Thus, the `alpha`-particle reverses its motion without ever actually touching the gold nucleus.
Promotional Banner

Topper's Solved these Questions

Similar Questions

Explore conceptually related problems

In a geiger - marsden experiment. Find the distance of closest approach to the nucleus of a 7.7 me v alpha - particle before it comes momentarily to rest and reverses its direction. (z for gold nucleus = 79) .

In a Geiger Marsden experiment, calculate the distance of closest approach to the nucleus of Z=80, when an alpha -particle of 8 MeV energy impings om it before it comes momentarily to rest and reverses its direction.

In a Geiger - Marsden experiment, calculate the distance of closest approach to the nucleus of Z=75, when an alpha - particle of 5 Me V enery impinges on it before it comes momentarily to rest and reverses its direction. How will the distance of closest approach be affected when the kinetic energy of the alpha -particle is doubled ?

In a Geiger- Marsden experiment, calculate the distance of closest approach to the nucleus of Z=80, when an alpha particle of 8 Me V energy impinges on it before it comes momentarily to rest and reverses its direction. How will the distance of closest approach be affected when the kinetic energy of the alpha- particle is doubled ?

Calculate the energy of an alpha -particle whose distance of closest approach with the gold nucleus is 29.5 fermi.

In Rutherford experiment, a 5.3 MeV alpha -particle moves towards the gold nucleus (Z=79). How close does the alpha -particle to get the centre of the nucleus, before it comes momentarily to rest and reverses its motion? ( epsi_(0)=8.8xx10^(-12)F//m )

Calculate the distance of closest approach when a proton of energy 3 MeV approaches a gold nucleus (Z=79).

What is the distance of closest approach to the nucleus of an alpha partile which undergoes scattering by 180^(@) is geiger-marsden experiment?

NCERT-ATOMS-Exercise
  1. In the original experiment, Geiger and Marsden calculated the distance...

    Text Solution

    |

  2. Choose the correct alternative form the clues given at the end of each...

    Text Solution

    |

  3. Suppose you are given a chance to repeat the alpha-particle scattering...

    Text Solution

    |

  4. What is the shortest wavelength present in the Paschen series of spect...

    Text Solution

    |

  5. A difference of 2.3 eV separates two energy levels in an atom. What is...

    Text Solution

    |

  6. The energy of the electron in the ground state of hydrogen atom is -13...

    Text Solution

    |

  7. A hydrogen atom initially in the ground level absorbs a photon, Which ...

    Text Solution

    |

  8. (a) Using the Bohr's model, calculate the speed of the electron in a h...

    Text Solution

    |

  9. The radius of innermost electron orbit of a hydrogen atom is 5.3xx10^(...

    Text Solution

    |

  10. A 12.5eV electron beam is used to bombard gaseous hydrogen at room tem...

    Text Solution

    |

  11. In accordance with the Bohr's model, find the quantum number that char...

    Text Solution

    |

  12. Answer the following questions, which help you understand the differen...

    Text Solution

    |

  13. The gravitational attraction between electron and proton in a hydrogen...

    Text Solution

    |

  14. Obtain an expression for the frequency of radiations emitted when a hy...

    Text Solution

    |

  15. Classically, an electron can be in any orbit around the nucleus of an ...

    Text Solution

    |

  16. The total energy of an electron in the first excited state of hydrogen...

    Text Solution

    |

  17. If Bohr’s quantization postulate (angular momentum = nh//2pi ) is a ba...

    Text Solution

    |

  18. Obtain the first Bohr radius and the ground state energy of a muonic h...

    Text Solution

    |