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(A) most of the mass of the atom is conc...

(A) most of the mass of the atom is concentrated in its nucleus.
(R) all alpha particles striking a gold sheet are scattered in different directions.

A

Both 'A' and 'R' are true and 'R' is the correct explanation of 'A'.

B

Both 'A' and 'R' are true and 'R' is not the correct explanation of 'A'

C

'A' is true and 'R' is false

D

'A' is false and 'R' is true

Text Solution

Verified by Experts

The correct Answer is:
C
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All the mass of an atom is concentrated in its-

Assertion:The whole mass of the atom is concentrated in the nucleus. Reason:The whole mass of the atom is concentrated in the nucleus Reason:The mass of a nucleus can be either less than or more than the sum of the masses of nucleons present in it.

Rutherford's alpha -particle scattering experiment showed that (i) electrons have negative charge. (ii) the mass and positive charge of the atom is concentrated in the nucleus. (iii) neutron exists in the nucleus. (iv) most of the space in atom is empty. Which of the above statements are correct ?

What is the ratio of mass of a gold atom and an alpha -particle ?

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) .

Rutherford model: The approximate size of the nucleus can be calculated by using energy conservation theorem in Rutherford's alpha -scattering experiment. If an alpha -particle is projected from infinity with speed v towards the nucleus having Z protons, then the alpha -particle which is reflected back or which is deflected by 180^@ must have approached closest to the nucleus .It can be approximated that alpha particle collides with the nucleus and gets back. Now if we apply the energy conservation equation at initial point and collision point then: (P.E.)_i= 0 , since P.E. of two charge system separated by infinite distance is zero. Finally the particle stops and then starts coming back. 1/2m_alpha v_alpha^2+0=0+(Kq_1q_2)/Rimplies 1/2m_alphav_alpha^2=K(2exxZe)/R implies R=(4KZe^2)/(m_alphav_alpha^2) Thus the radius of nucleus can be calculated using above equation. The nucleus is so small a particle that we can't define a sharp boundary for it Radius of a particular nucleus is calculated by the projection of alpha -particle from infinity at a particular speed. Let this radius is the true radius . If the radius calculation for the same nucleus is made by another alpha -particle with half of the earlier speed, then the percentage error involved in the radius calculation is :

What is the distance of closest approach to the nucleus for an alpha -particle of energy 5 MeV which undergoes scattering in the Gieger-Marsden experiment.

Heavy radioactive nucleus decay through alpha- decay also. Consider a radioactive nucleus x. It spontaneously undergoes decay at rest resulting in the formation of a daughter nucleus y and the emission of an alpha- particle. The radioactive reaction can be given by x rarr y + alpha However , the nucleus y and the alpha- particle will be in motion. Then a natural question arises that what provides kinetic energy to the radioactive products. In fact, the difference of masses of the decaying nucleus and the decay products provides for the energy that is shared by the daughter nucleus and the alpha- particle as kinetic energy . We know that Einstein's mass-energy equivalence relation E = m c^(2) . Let m_(x), m_(y) and m_(alpha) be the masses of the parent nucleus x, the daughter nucleus y and alpha- particle respectively. Also the kinetic energy of alpha- particle just after the decay is E_(0) . Assuming all motion of to be non-relativistic. Which of the following is correct ?

All nuclei consist of two types of particles- protaon and neutrons. Nuclear force is the strongest froce. Stability of nucleus is determined by the neutron - proton ratio or mass defect or binding energy per nucleus or packing fraction. Shape of nucleus is calculated by quadrupole moment. Spin of nucleus depends on even or odd mass number. Volume of nucleus depends on the mass number. Whole mass of the atom (nearly 99 % ) is centered at the nucleus. Magnetic moment of the nucleus is measured in terms of the nuclear magnetons. Volume (V) of the nucleus is related to mass number (A) as

Thickness of the foil of gold used in alpha -particle scattering experiment is

AAKASH SERIES-ATOMS-EXERCISE -IB
  1. When an electron of the hydrogen atom jumps from a higher to lower en...

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  2. (A) hydrogen atom consists of only one electron but its emission spect...

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  3. (A) for the scattering of alpha- particles at large angles, only the n...

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  4. Statement-1 : Between any two given energy levels, the number of abosr...

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  5. (A) atoms of each element are stable and emit characteristic spectrum....

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  6. (A) most of the mass of the atom is concentrated in its nucleus. (R)...

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  7. (A) bohr's third postulaate states that the stationary orbits are thos...

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  8. (A) the total energy of an electron revolving in any stationary orbit ...

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  9. (A) in the experiment of alpha particle scattering, extremely thin gol...

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  10. (A) in alpha particle scattering number of alpha paritcle undergoing h...

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  11. (A) electrons int the atom are held due to coulomb forces. (R) the a...

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  12. (A) according to classical electromagnetic theory an accelerated parti...

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  13. Assertion : A laser beam of 0.2 watt power can drill holes through a m...

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  14. Assertion (A) : LASER is used to measure distant object as moon. Re...

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  15. Assertion In He-Nelaser, population inversion takes place between ener...

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  16. Assertion : Crystalline solids can cause Xrays to diffract. Reason...

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  17. A: In beta-decay an electron is emitted by the nucleus R: Electrons ...

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  18. Assertion : X-ray astronomy is possible only from satellites orbiting ...

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  19. According to Bohr's atomic model

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  20. Assertion : Total energy of electron in an hydrogen atom is negative. ...

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