CBSE Class 12 Physics Chapter 12 Atom NCERT Solutions

(a) The average angle of deflection predicted by Thomson's model is considered to be about the same as that predicted by Rutherford's model. Both models, in their own ways, account for an average deflection, although the underlying physics and the magnitude of deflections differ significantly in their detailed predictions for individual particles.

(b) The probability of backward scattering (angles greater than 90°) predicted by Thomson's model is much less than that predicted by Rutherford's model. This is a key distinction, as Rutherford's nuclear model explained the occasional large-angle scattering observed experimentally, which Thomson's model, with its uniformly distributed positive charge, could not adequately explain.

(c) The experimental observation that the number of a-particles scattered at moderate angles is proportional to the thickness 't' (for small t) provides a crucial clue: it suggests that the scattering process is predominantly due to single collisions. If multiple scattering were dominant, the dependence on thickness would be more complex than a simple linear relationship.

(d) It is completely wrong to ignore multiple scattering for the calculation of the average angle of scattering of a-particles by a thin foil in Thomson's model. In Thomson's model, the positive charge and mass are assumed to be spread out, leading to very small deflections in a single collision. Therefore, to achieve the observed average scattering angles, the cumulative effect of multiple small-angle deflections (multiple scattering) must be considered.

The radius of the first Bohr orbit ($r$) in a hydrogen atom, when bound by electrostatic (Coulomb) force, is given by the formula:

$$r\; =\; \backslash frac\{4\backslash pi\; \backslash epsilon\_0\; \backslash hbar^2\}\{m\_e\; m\_e\; e^2\}$$

where $\backslash epsilon\_0$ is the permittivity of free space, $\backslash hbar\; =\; \backslash frac\{h\}\{2\backslash pi\}$ is the reduced Planck constant, $m\_e$ is the mass of the electron, and $e$ is the magnitude of the electron's charge.

Let's consider the case where the electron and proton are bound by gravitational attraction instead of Coulomb attraction. The gravitational force ($F\_G$) between the electron (mass $m\_e$) and the proton (mass $m\_p$) at a distance $r$ is given by:

$$F\_G\; =\; \backslash frac\{G\; m\_p\; m\_e\}\{r^2\}$$

where $G$ is the gravitational constant.

For a stable orbit, this gravitational force must provide the necessary centripetal force ($F\_c$) for the electron moving with velocity $v$ around the proton:

$$F\_G\; =\; F\_c\; =\; \backslash frac\{m\_e\; v^2\}\{r\}$$

So, $\backslash frac\{G\; m\_p\; m\_e\}\{r^2\}\; =\; \backslash frac\{m\_e\; v^2\}\{r\}$, which implies $v^2\; =\; \backslash frac\{G\; m\_p\}\{r\}$.

According to Bohr's quantization condition, the angular momentum ($L$) of the electron is quantized:

$$L\; =\; m\_e\; v\; r\; =\; n\; \backslash hbar$$

For the first Bohr orbit, $n=1$, so $m\_e\; v\; r\; =\; \backslash hbar$, which gives $v\; =\; \backslash frac\{\backslash hbar\}\{m\_e\; r\}$.

Substituting this expression for $v$ into the equation for $v^2$ from the force balance:

$$\backslash left(\backslash frac\{\backslash hbar\}\{m\_e\; r\}\backslash right)^2\; =\; \backslash frac\{G\; m\_p\}\{r\}$$

$$\backslash frac\{\backslash hbar^2\}\{m\_e^2\; r^2\}\; =\; \backslash frac\{G\; m\_p\}\{r\}$$

Solving for $r$ (the radius of the first Bohr orbit under gravitational force):

$$r\_\{grav\}\; =\; \backslash frac\{\backslash hbar^2\}\{G\; m\_p\; m\_e^2\}$$

Now, let's compare this with the radius of the first Bohr orbit under Coulomb force ($r\_\{coulomb\}$):

$$r\_\{coulomb\}\; =\; \backslash frac\{4\backslash pi\; \backslash epsilon\_0\; \backslash hbar^2\}\{m\_e\; e^2\}$$

The ratio of the radii is:

$$\backslash frac\{r\_\{grav\}\}\{r\_\{coulomb\}\}\; =\; \backslash frac\{\backslash hbar^2\; /\; (G\; m\_p\; m\_e^2)\}\{4\backslash pi\; \backslash epsilon\_0\; \backslash hbar^2\; /\; (m\_e\; e^2)\}\; =\; \backslash frac\{m\_e\; e^2\}\{G\; m\_p\; m\_e^2\; 4\backslash pi\; \backslash epsilon\_0\}\; =\; \backslash frac\{e^2\}\{4\backslash pi\; \backslash epsilon\_0\; G\; m\_p\; m\_e\}$$

We know that the Coulomb force is $F\_C\; =\; \backslash frac\{e^2\}\{4\backslash pi\; \backslash epsilon\_0\; r^2\}$ and the gravitational force is $F\_G\; =\; \backslash frac\{G\; m\_p\; m\_e\}\{r^2\}$. The ratio of these forces is $\backslash frac\{F\_C\}\{F\_G\}\; =\; \backslash frac\{e^2\; /\; (4\backslash pi\; \backslash epsilon\_0\; r^2)\}\{G\; m\_p\; m\_e\; /\; r^2\}\; =\; \backslash frac\{e^2\}\{4\backslash pi\; \backslash epsilon\_0\; G\; m\_p\; m\_e\}$.

The problem states that the gravitational attraction is weaker than the Coulomb attraction by a factor of about $10^\{-40\}$, which means $F\_G\; =\; 10^\{-40\}\; F\_C$, or $\backslash frac\{F\_C\}\{F\_G\}\; \backslash approx\; 10^\{40\}$.

Therefore, $\backslash frac\{r\_\{grav\}\}\{r\_\{coulomb\}\}\; \backslash approx\; 10^\{40\}$.

This implies that the radius of the first Bohr orbit if bound by gravitational attraction would be approximately $10^\{40\}$ times larger than the actual Bohr orbit radius. This is an astronomically large increase, highlighting the immense difference in strength between the electromagnetic and gravitational forces at the atomic scale.