Atomic Models and Scattering of Alpha particle Questions in English

(C) Rutherford's alpha-particle scattering experiment led to the discovery of the nucleus. His atomic model proposed that the entire positive charge and most of the mass of an atom are concentrated in a very small central region called the nucleus. Therefore,it could successfully account for the positively charged central core of an atom. It failed to explain the stability of atoms and the origin of line spectra,which were later addressed by Bohr's model.

(B) The distance of closest approach $(d)$ is the distance at which the kinetic energy of the $\alpha$-particle is completely converted into electrostatic potential energy.
$d = \frac{1}{4 \pi \varepsilon_{0}} \frac{(Z_1 e)(Z_2 e)}{K}$
Here,$Z_1 = 2$ (for $\alpha$-particle),$Z_2 = 79$ (for gold nucleus),and $K = 5 \text{ MeV} = 5 \times 10^6 \times 1.6 \times 10^{-19} \text{ J} = 8 \times 10^{-13} \text{ J}$.
Substituting the values:
$d = (9 \times 10^9) \times \frac{(2 \times 1.6 \times 10^{-19}) \times (79 \times 1.6 \times 10^{-19})}{8 \times 10^{-13}}$
$d = \frac{9 \times 10^9 \times 2 \times 79 \times 2.56 \times 10^{-38}}{8 \times 10^{-13}}$
$d \approx 4.55 \times 10^{-14} \text{ m} = 4.55 \times 10^{-12} \text{ cm}$.
Thus,the order of magnitude is $10^{-12} \text{ cm}$.

(B) Ernest Rutherford is the scientist credited with the discovery of the atomic nucleus.
In $1911$,he conducted the famous alpha-particle scattering experiment (Gold Foil Experiment).
He observed that a small fraction of alpha particles were deflected by large angles,which led him to conclude that the positive charge and most of the mass of the atom are concentrated in a very small central region called the nucleus.

(C) At a sufficient distance,the electric potential energy is zero. So,$U_1 = 0$.
At position $(1)$,the kinetic energy is $K_1 = \frac{1}{2} mv^2 = \frac{P^2}{2m}$.
At the distance of closest approach $(2)$,the kinetic energy is $K_2 = 0$.
The electric potential energy at distance $d_c$ is $U_2 = \frac{1}{4 \pi \epsilon_0} \frac{q_\alpha q_n}{d_c}$.
By the law of conservation of energy:
$U_1 + K_1 = U_2 + K_2$
$0 + \frac{P^2}{2m} = \frac{1}{4 \pi \epsilon_0} \frac{q_\alpha q_n}{d_c} + 0$
From this,we see that the distance of closest approach $d_c \propto \frac{1}{P^2}$.
Therefore,$\frac{d_2}{d_1} = \frac{P_1^2}{P_2^2}$.
Given $d_1 = d$,$P_1 = P$,and $P_2 = 1.5 P = \frac{3}{2} P$:
$d_2 = d \times \frac{P^2}{(1.5 P)^2} = d \times \frac{P^2}{2.25 P^2} = d \times \frac{1}{2.25} = d \times \frac{4}{9} = \frac{4d}{9}$.

(D) At the distance of closest approach,the entire kinetic energy of the alpha particle is converted into electrostatic potential energy.
$K.E. = U$
$K \times 10^6 \times 1.6 \times 10^{-19} \text{ J} = \frac{1}{4 \pi \epsilon_0} \frac{(2e)(Ze)}{r}$
Substituting the value of $\frac{1}{4 \pi \epsilon_0} = 9 \times 10^9 \text{ N m}^2/\text{C}^2$ and $e = 1.6 \times 10^{-19} \text{ C}$:
$r = \frac{2 \times 9 \times 10^9 \times Z \times (1.6 \times 10^{-19})^2}{K \times 10^6 \times 1.6 \times 10^{-19}}$
$r = \frac{18 \times 10^9 \times 1.6 \times 10^{-19} \times Z}{K \times 10^6}$
$r = \frac{28.8 \times 10^{-10}}{K \times 10^6} Z$
$r = 28.8 \times 10^{-16} \frac{Z}{K} \text{ m}$

(A) The impact parameter $b$ is given by the formula: $b = \frac{Z e^2 \cot(\theta/2)}{4 \pi \varepsilon_0 (\frac{1}{2} m v^2)}$.
For scattering at an angle $\theta = 180^{\circ}$,we have $\cot(180^{\circ}/2) = \cot(90^{\circ}) = 0$.
Substituting this into the formula,we get $b = 0$.
$A$ zero impact parameter implies that the $\alpha$-particle is directed straight towards the centre of the nucleus,leading to a head-on collision and backscattering at $180^{\circ}$.
Therefore,both $A$ and $R$ are true and $R$ is the correct explanation for $A$.

(C) The Franck-Hertz experiment demonstrated the existence of discrete energy levels in atoms by bombarding mercury vapor with electrons. When the kinetic energy of the electrons reached a specific threshold,they lost energy to the mercury atoms,causing them to transition to higher energy states. This confirms the prediction of quantum theory that the energy states of an atom are quantised.

(B) In a hydrogen atom,the potential energy $(PE)$ of an electron in the $n$th orbit is given by $PE = -\frac{kZe^2}{r}$.
For higher values of $n$,the radius $r$ increases because $r \propto n^2$.
As $r$ increases,the magnitude of the negative potential energy decreases,meaning the value of the potential energy increases (becomes less negative).
Thus,statement $(A)$ is correct.
In Thomson's model,an atom is described as a spherical cloud of positive charge with electrons embedded within it,similar to a plum pudding. Thus,statement $(B)$ is correct.
According to Bohr's model,an electron in an atom is located at a definite distance from the nucleus and revolves with a definite velocity. However,Heisenberg's uncertainty principle states that it is impossible to determine the exact position and momentum of a moving electron simultaneously.
Therefore,Bohr's model contradicts the uncertainty principle,making statement $(C)$ incorrect.
Hence,option $(b)$ is the correct answer.

(A) At the distance of closest approach $(r)$,the entire kinetic energy of the $\alpha$-particle is converted into electrostatic potential energy.
By the principle of conservation of mechanical energy:
$K.E._i + P.E._i = K.E._f + P.E._f$
Given $K.E._i = 7.9 \text{ MeV} = 7.9 \times 10^6 \times 1.6 \times 10^{-19} \text{ J}$,$P.E._i = 0$,$K.E._f = 0$,and $P.E._f = \frac{1}{4 \pi \epsilon_0} \frac{(2e)(Ze)}{r}$.
$7.9 \times 10^6 \times 1.6 \times 10^{-19} = \frac{9 \times 10^9 \times 2 \times 79 \times (1.6 \times 10^{-19})^2}{r}$
Solving for $r$:
$r = \frac{9 \times 10^9 \times 2 \times 79 \times 1.6 \times 10^{-19}}{7.9 \times 10^6} = 2.88 \times 10^{-14} \text{ m}$
The diameter of the nucleus is $D = 2r = 2 \times 2.88 \times 10^{-14} = 5.76 \times 10^{-14} \text{ m}$.

(A) The distance of closest approach $r_0$ is the distance where the kinetic energy of the alpha particle is completely converted into electrostatic potential energy.
Using the principle of conservation of energy:
$K_i + U_i = K_f + U_f$
At the distance of closest approach,final kinetic energy $K_f = 0$.
$K_i = \frac{1}{4 \pi \epsilon_0} \cdot \frac{(Ze)(2e)}{r_0}$
Given:
$K_i = 7.7 \text{ MeV} = 7.7 \times 10^6 \times 1.6 \times 10^{-19} \text{ J}$
$Z = 79$ (for gold)
$e = 1.6 \times 10^{-19} \text{ C}$
$\frac{1}{4 \pi \epsilon_0} = 9 \times 10^9 \text{ N m}^2/\text{C}^2$
Substituting the values:
$7.7 \times 10^6 \times 1.6 \times 10^{-19} = \frac{9 \times 10^9 \times (79 \times 1.6 \times 10^{-19}) \times (2 \times 1.6 \times 10^{-19})}{r_0}$
$r_0 = \frac{9 \times 10^9 \times 79 \times 2 \times 1.6 \times 10^{-19}}{7.7 \times 10^6}$
$r_0 \approx 2.95 \times 10^{-14} \text{ m}$

(D) In Rutherford's alpha-particle scattering experiment,alpha particles are positively charged and experience a strong electrostatic repulsive force when they approach the gold nucleus.
Most of the atom is empty space,allowing the majority of alpha particles to pass through undeflected.
$A$ small fraction of alpha particles rebound because they undergo a near head-on collision with the dense,positively charged nucleus.
Since the nucleus occupies a very small volume compared to the total volume of the atom,the probability of such a collision is extremely low.
Thus,the rebounding is primarily due to the head-on collision with the tiny,massive nucleus.