CBSE Class 12 Physics Chapter 22: Atoms NCERT Solutions

Solution:

1. The average angle of deflection of a-particles by a thin gold foil predicted by Thomson's model is about the same as that predicted by Rutherford's model. Both models consider the interaction between the alpha-particle and the charged constituents of the atom, leading to a comparable average deflection, although the mechanisms differ.

2. The probability of scattering of a-particles at angles greater than 90° predicted by Thomson's model is much less than that predicted by Rutherford's model. This is because Thomson's model, with its diffuse positive charge, is less likely to cause a strong repulsive force needed for large-angle scattering compared to Rutherford's model with its concentrated, positively charged nucleus.

3. The experimental observation that the number of a-particles scattered at moderate angles is proportional to the thickness 't' of the foil (for small t) provides a crucial clue. It suggests that the scattering process is cumulative. Each atom in the foil has a certain probability of scattering an alpha-particle. As the thickness increases, the number of atoms encountered by the alpha-particle increases linearly, leading to a proportional increase in the probability of scattering. This implies that scattering is primarily due to single collisions, and the total scattering effect is the sum of effects from individual atoms.

4. 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 is spread out, so a single collision with an atom causes only a very small deflection. To achieve the significant average scattering angles observed experimentally, one must consider the cumulative effect of multiple small-angle scatterings as the alpha-particle passes through the foil. In contrast, Rutherford's model, with its concentrated nucleus, predicts that large-angle scattering can occur from a single strong interaction, making the assumption of single scattering more plausible for estimating average angles in certain scenarios.

Solution:

The radius of the first Bohr orbit in a hydrogen atom is determined by the balance between the Coulomb attraction and the kinetic energy of the electron. The formula for the radius of the n-th Bohr orbit is generally given by:

$$r\_n\; =\; \backslash frac\{n^2\; \backslash hbar^2\}\{m\_e\; k\; e^2\}$$

where $\backslash hbar\; =\; \backslash frac\{h\}\{2\backslash pi\}$ is the reduced Planck constant, $m\_e$ is the mass of the electron, $k\; =\; \backslash frac\{1\}\{4\backslash pi\; \backslash epsilon\_0\}$ is Coulomb's constant, $e$ is the elementary charge, and $n$ is the principal quantum number.

For the first Bohr orbit, $n=1$, so the radius $r\_1$ is:

$$r\_1\; =\; \backslash frac\{\backslash hbar^2\}\{m\_e\; k\; e^2\}\; =\; \backslash frac\{1\}\{4\backslash pi\; \backslash epsilon\_0\}\; \backslash frac\{\backslash hbar^2\}\{m\_e\; e^2\}$$

Let's calculate the values of the constants:

$\backslash epsilon\_0$ = Permittivity of free space $\approx \; 8.854\; \backslash times\; 10^\{-12\}\; \backslash ,\; C^2\; N^\{-1\}\; m^\{-2\}$

h = Planck's constant $\approx \; 6.63\; \backslash times\; 10^\{-34\}\; \backslash ,\; Js$

$\backslash hbar\; =\; \backslash frac\{h\}\{2\backslash pi\}\; \approx \; 1.055\; \backslash times\; 10^\{-34\}\; \backslash ,\; Js$

$m\_e$ = Mass of an electron $\approx \; 9.1\; \backslash times\; 10^\{-31\}\; \backslash ,\; kg$

$e$ = Charge of an electron $\approx \; 1.602\; \backslash times\; 10^\{-19\}\; \backslash ,\; C$

$m\_p$ = Mass of a proton $\approx \; 1.67\; \backslash times\; 10^\{-27\}\; \backslash ,\; kg$

The Coulomb force of attraction between an electron and a proton is:

$$F\_C\; =\; \backslash frac\{e^2\}\{4\backslash pi\; \backslash epsilon\_0\; r^2\}$$

The gravitational force of attraction between an electron and a proton is:

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

Where $G$ is the gravitational constant, $G\; \approx \; 6.674\; \backslash times\; 10^\{-11\}\; \backslash ,\; N\; m^2\; kg^\{-2\}$.

If the electron and proton were bound by gravitational attraction, the radius of the first Bohr orbit ($r\_G$) would be determined by balancing the gravitational force with the required centripetal force (or related quantum condition). Using a similar approach as for the Coulomb force, we can equate the gravitational force to the force required to maintain a specific orbit. A simplified approach is to consider the condition for the first Bohr orbit where angular momentum $L\; =\; m\_e\; v\; r\; =\; \backslash hbar$. The kinetic energy $K\; =\; \backslash frac\{1\}\{2\}\; m\_e\; v^2\; =\; \backslash frac\{1\}\{2\}\; \backslash frac\{\backslash hbar^2\}\{m\_e\; r^2\}$. The potential energy is $U\; =\; -\backslash frac\{G\; m\_p\; m\_e\}\{r\}$. For a stable orbit, the total energy $E\; =\; K\; +\; U$ is related to the force. A more direct way is to find the radius $r\_G$ where the gravitational force provides the necessary binding.

Let's find the radius $r\_G$ for the first Bohr orbit under gravitational force. The condition for the first Bohr orbit is often derived by setting the electrostatic force equal to $m\_e\; v^2\; /\; r$ and using $L\; =\; m\_e\; v\; r\; =\; \backslash hbar$. If we replace the electrostatic force with the gravitational force, we get:

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

And $m\_e\; v\; r\_G\; =\; \backslash hbar\; \backslash implies\; v\; =\; \backslash frac\{\backslash hbar\}\{m\_e\; r\_G\}$.

Substituting $v$ into the force equation:

$$\backslash frac\{G\; m\_p\; m\_e\}\{r\_G^2\}\; =\; \backslash frac\{m\_e\}\{r\_G\}\; \backslash left(\backslash frac\{\backslash hbar\}\{m\_e\; r\_G\}\backslash right)^2\; =\; \backslash frac\{\backslash hbar^2\}\{m\_e\; r\_G^3\}$$

Solving for $r\_G$:

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

Now, let's substitute the values:

$r\_G\; =\; \backslash frac\{(1.055\; \backslash times\; 10^\{-34\}\; \backslash ,\; Js)^2\}\{(6.674\; \backslash times\; 10^\{-11\}\; \backslash ,\; N\; m^2\; kg^\{-2\})\; (1.67\; \backslash times\; 10^\{-27\}\; \backslash ,\; kg)\; (9.1\; \backslash times\; 10^\{-31\}\; \backslash ,\; kg)^2\}$

$r\_G\; =\; \backslash frac\{1.113\; \backslash times\; 10^\{-68\}\}\{6.674\; \backslash times\; 10^\{-11\}\; \backslash times\; 1.67\; \backslash times\; 10^\{-27\}\; \backslash times\; 8.281\; \backslash times\; 10^\{-61\}\}$

$r\_G\; =\; \backslash frac\{1.113\; \backslash times\; 10^\{-68\}\}\{9.27\; \backslash times\; 10^\{-98\}\}\; \approx \; 1.2\; \backslash times\; 10^\{30\}\; \backslash ,\; m$

The actual radius of the first Bohr orbit ($r\_1$) is approximately $0.53\; \backslash times\; 10^\{-10\}\; \backslash ,\; m$.

Comparing $r\_G$ with $r\_1$:

$\backslash frac\{r\_G\}\{r\_1\}\; =\; \backslash frac\{1.2\; \backslash times\; 10^\{30\}\; \backslash ,\; m\}\{0.53\; \backslash times\; 10^\{-10\}\; \backslash ,\; m\}\; \approx \; 2.26\; \backslash times\; 10^\{40\}$

This result shows that if the electron and proton were bound by gravitational attraction, the radius of the first Bohr orbit would be enormously larger (by a factor of about $10^\{40\}$) than the actual radius determined by Coulomb attraction. This highlights how incredibly weak gravity is compared to the electromagnetic force at the atomic scale.