CBSE Class 12 Physics Chapter 23: Atoms NCERT Solutions

1. The sizes of the atoms in Thomson's model and Rutherford's model are generally considered to be of the same order of magnitude, approximately $10^\{-10\}$ m. Therefore, the size is no different from.
2. In the ground state of Thomson's model, the electrons are considered to be in stable equilibrium due to the distribution of positive charge. However, in Rutherford's model, electrons orbiting the nucleus are constantly accelerating and thus experience a net force towards the nucleus, and classically, they should radiate energy and spiral inwards.
3. A classical atom based on Rutherford's model is doomed to collapse because the orbiting electrons, according to classical electromagnetism, should lose energy and fall into the nucleus.
4. An atom has a nearly continuous mass distribution in Thomson's model, where the positive charge and mass are spread throughout the volume. In contrast, Rutherford's model has a highly non-uniform mass distribution because most of the mass is concentrated in the tiny nucleus.
5. The positively charged part of the atom possesses most of the mass in both the models. In Thomson's model, the positive charge (and associated mass) is spread throughout the atom. In Rutherford's model, the positive charge and almost all the mass are concentrated in the nucleus.

If the alpha-particle scattering experiment were repeated using a thin sheet of solid hydrogen instead of a gold foil, the results would be significantly different. The alpha-particles (which are helium nuclei, $^\{4\}\_\{2\}\backslash text\{He\}$) have a mass of approximately $4\; \backslash text\{\; amu\}$ and are incident at high speeds. The nucleus of hydrogen ($^\{1\}\_\{1\}\backslash text\{H\}$) has a mass of approximately $1\; \backslash text\{\; amu\}$. In a collision, the scattering angle depends on the relative masses of the colliding particles. When a lighter nucleus (hydrogen) is the target and a heavier particle (alpha-particle) is the projectile, the alpha-particle will not be deflected by large angles. The interaction would be more like a glancing collision, and the alpha-particle would continue with only a slight change in direction. Significant backscattering, as observed with gold nuclei (which are much heavier than alpha-particles), would not be expected. The number of alpha-particles scattered at large angles would be much smaller compared to the experiment with gold foil.

The Paschen series in the hydrogen spectrum corresponds to electron transitions from higher energy levels to the principal quantum number $n\_1\; =\; 3$. The wavelength ($\backslash lambda$) of the emitted spectral lines can be calculated using the Rydberg formula:

$$\backslash frac\{1\}\{\backslash lambda\}\; =\; R\; \backslash left(\; \backslash frac\{1\}\{n\_1^2\}\; -\; \backslash frac\{1\}\{n\_2^2\}\; \backslash right)$$

where $R$ is the Rydberg constant ($R\; \backslash approx\; 1.097\; \backslash times\; 10^7\; \backslash text\{\; m\}^\{-1\}$), $n\_1$ is the lower energy level, and $n\_2$ is the higher energy level ($n\_2\; >\; n\_1$).

For the Paschen series, $n\_1\; =\; 3$. The shortest wavelength occurs when the energy difference between the levels is maximum, which corresponds to the transition from the highest possible energy level ($n\_2\; =\; \backslash infty$) to the lowest level of the series ($n\_1\; =\; 3$).

Substituting these values into the Rydberg formula:

$$\backslash frac\{1\}\{\backslash lambda\_\{\backslash text\{min\}\}\}\; =\; R\; \backslash left(\; \backslash frac\{1\}\{3^2\}\; -\; \backslash frac\{1\}\{\backslash infty^2\}\; \backslash right)\; =\; R\; \backslash left(\; \backslash frac\{1\}\{9\}\; -\; 0\; \backslash right)\; =\; \backslash frac\{R\}\{9\}$$

Therefore, the shortest wavelength is:

$$\backslash lambda\_\{\backslash text\{min\}\}\; =\; \backslash frac\{9\}\{R\}$$

Using the value of the Rydberg constant $R\; =\; 1.097\; \backslash times\; 10^7\; \backslash text\{\; m\}^\{-1\}$:

$$\backslash lambda\_\{\backslash text\{min\}\}\; =\; \backslash frac\{9\}\{1.097\; \backslash times\; 10^7\; \backslash text\{\; m\}^\{-1\}\}\; \backslash approx\; 8.189\; \backslash times\; 10^\{-7\}\; \backslash text\{\; m\}$$

Converting this to nanometers (1 nm = $10^\{-9\}$ m):

$$\backslash lambda\_\{\backslash text\{min\}\}\; \backslash approx\; 8.189\; \backslash times\; 10^\{-7\}\; \backslash text\{\; m\}\; \backslash times\; \backslash frac\{10^9\; \backslash text\{\; nm\}\}\{1\; \backslash text\{\; m\}\}\; \backslash approx\; 818.9\; \backslash text\{\; nm\}$$

The shortest wavelength present in the Paschen series of spectral lines is approximately 818.9 nm.