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18 Questions around this concept.
An particle of energy is scattered through by a fixed uranium nucleus. The distance of the closest approach is of the order of
An particle of energy is scattered through by a gold nucleus. The distance of the closest approach is of the order of:
The distance of the closest approach of an -particle fired towards a nucleus with momentum , is . If the momentum of the - particle is then the corresponding distance of the closest approach is:
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A beam of fast moving alpha particles were directed towards a thin film of gold. The parts A',B' and C' of the transmitted and reflected beams corresponding to the incident parts A, B and C of the beam, are shown in the adjoining diagram. The number of alpha particles in:
Rutherford's model of atom (I)
J.J Thomson's model - J. J. Thomson, who discovered the electron in 1897, proposed the plum pudding model of the atom in 1904 before the discovery of the atomic nucleus in order to include the electron in the atomic model. In Thomson’s model, the atom is composed of electrons surrounded by a soup of positive charge to balance the electrons’ negative charges, like negatively charged “plums” surrounded by positively charged “pudding”.
The Rutherford Model (Gold Foil Experiment ) :
Rutherford and his collegues Geiger and Marsden bombarded a thin gold foil of thickness approximately $8.6 \times 10^{-6}$ cm with a beam of alpha particles in vacuum. They used gold since it is highly malleable, producing sheets that can be only a few atoms thick, thereby ensuring smooth passage of the alpha particles. A circular screen coated with zinc sulphide surrounded the foil. Since the positively charged alpha particles possess mass and move very fast, it was hypothesized that they would penetrate the thin gold foil and land themselves on the screen, producing fluorescence in the part they struck.
In line with the plum pudding model, since the positive charge of atoms was evenly distributed and too small as compared to that of the alpha particles, the deflection of the particulate matter, if any, was predicted to be less than a small fraction of a degree.
Observations :
Conclusion: A highly concentraíted positive charge at the center of an atom that caused an electrostatic repulsion of the particles strong enough to bounce them back to their source. The particles that got deflected by huge angles passed close to the said concentrated mass. Most of the particles passed undeviated as there was no obstruction to their path, proving that the majority of an atom is empty. Rutherford drew the conclusion that since the dense alpha particles could be deflected by the central core, it shows that almost the entire mass of the atom is concentrated there. Rutherford named it the “nucleus” after performing the experiment in various gases.
Rutherford scattering formula
For a detector at a specific angle $(\theta)$ with respect to the incident beam, the number of particles per unit area striking the detector is given by the Rutherford formula:
$$
N(\theta) \propto \frac{1}{\sin ^4\left(\frac{\theta}{2}\right)}
$$
Rutherford's model of atom (II)
Distance of closest approach :
The minimum distance from the nucleus up to which the $\alpha$ - particle approach, is called the distance of closest approach $\left(r_0\right)$. At this distance the entire initial kinetic energy has been converted into potential energy so
$
\begin{aligned}
(\text { K.E. })_{\text {initial }}=K=\frac{1}{2} m v^2 & =\frac{1}{4 \pi \varepsilon_0} \cdot \frac{(Z e) 2 e}{r_0} \\
r_0 & =\frac{Z e^2}{m v^2 \pi \varepsilon_0}=\frac{4 k Z e^2}{m v^2}=\frac{2 k Z e^2}{K}
\end{aligned}
$
Impact parameter: It is defined as the perpendicular distance of the velocity of the alpha-particle from the centre of the nucleus when it is far away from the atom. The shape of the trajectory of the scattered alpha particle depends on the impact parameter ' $b$ ' and the nature of the potential field. Rutherford deduced the following relationship between the impact parameter ' b ' and the scattering angle $\theta$. It is given as
$
\begin{aligned}
b & =\frac{Z e^2 \cot (\theta / 2)}{4 \pi \varepsilon_0\left(\frac{1}{2} m v^2\right)} \\
\Rightarrow b & \propto \cot (\theta / 2)
\end{aligned}
$
Conclusion and drawback of Rutherford model:
Conclusion:
Limitations of Rutherford model:
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