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A protom of energy 1 MeV is incident hea...

A protom of energy 1 MeV is incident head-on on a gold nucleus (Z = 79), and is scattered through an angle of `180^(@)`. Calculate the distance of nearest approach.

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The correct Answer is:
`11.37 xx 10^(-4) m`
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SL ARORA-ATOMS, NUCLEI AND MOLECULES-Problems For Self Practice
  1. In a head-on collision between and alpha-particle and a gold nucleus, ...

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  2. An alpha particle of K.E. 10^(-12)J exhibits back scattering from a go...

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  3. A protom of energy 1 MeV is incident head-on on a gold nucleus (Z = 79...

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  4. In a head-on collision between an alpha-particle and a gold nucleus, t...

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  5. An alpha-particle of kinetic energy 7.68 MeV is projected towards the ...

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  6. An alpha particle of energy 4MeV is scattered through an angle of 180^...

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  7. What is the distance of closest approach to the nucleus for an alpha-p...

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  8. What is the impact parameter at which the scattering angle is 10^(@) f...

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  9. A 10kg satellite circles earth once every 2hr in an orbit having a rad...

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  10. At what speed must an electron revolve around the nucleus of hydrogen ...

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  11. The innermost orbit of the hydrogen atom has a diameter of 1.06Å what ...

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  12. Calculate the energy of the hydrogen atom in the states n = 4 and n = ...

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  13. Calculate (i) the radius of the second orbit of hydrogen atom, and (ii...

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  14. Calculate the frequency of the photon, which can excite the electron ...

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  15. The ground state energy of hydrogen atom is -13.6eV. If an electron ma...

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  16. Show that the ionisation potential of hydrogen atom is 13.6 volt.

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  17. Calculate the ionisation potential of a single ionised He-atom.

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  18. Calculate the minimum energy that must be given to a hydrogen atom so ...

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  19. What energy in eV will be required to excite the ground state electron...

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  20. By giving energy to a hydrogen atom in energy state n=1. If the potent...

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