CBSE Class 12 Physics Chapter 12 Atom 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 Thomson's model, electrons are embedded in a sphere of uniform positive charge, leading to a stable equilibrium. In Rutherford's model, electrons orbit the nucleus and are expected to radiate energy, experiencing a net force towards the nucleus. Thus, in the ground state of Thomson's model, electrons are in stable equilibrium, while in Rutherford's model, electrons always experience a net force.
3. According to classical electromagnetism, an accelerating charged particle (like an electron orbiting a nucleus) should continuously radiate energy and spiral into the nucleus. This leads to the collapse of the atom. Therefore, a classical atom based on Rutherford's model is doomed to collapse.
4. In Thomson's model, the positive charge and mass are assumed to be uniformly distributed throughout the atom's volume. In Rutherford's model, the mass and positive charge are concentrated in a tiny nucleus, making the mass distribution highly non-uniform.
5. In both Thomson's model (where the positive charge is spread out) and Rutherford's model (where the positive charge is concentrated in the nucleus), the positively charged part of the atom possesses most of the mass. Thus, the answer is both the models.

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 used in the experiment have a mass approximately 4 atomic mass units (amu) and are positively charged. Gold has a much larger atomic mass (approximately 197 amu).

Hydrogen, on the other hand, has the smallest atomic mass, with its nucleus (a single proton) having a mass of approximately 1 amu. When an alpha particle (mass $\backslash approx\; 4$ amu) collides with a hydrogen nucleus (mass $\backslash approx\; 1$ amu), the alpha particle is much heavier than the target nucleus.

In such a collision, the alpha particle would transfer only a small amount of momentum and energy to the much lighter hydrogen nucleus. Consequently, the deflection angles of the alpha particles would be very small. It would be highly unlikely for an alpha particle to be scattered backward, as observed with the gold foil, because the heavier alpha particle would not be significantly repelled or deflected by the much lighter proton.

Therefore, we would expect very few, if any, large-angle deflections, and certainly no backscattering, if solid hydrogen were used as the target material.

The Paschen series in the hydrogen spectrum corresponds to electron transitions from higher energy levels to the principal quantum level $n\_1\; =\; 3$. The wavelength ($\backslash lambda$) of the emitted spectral line is given by 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 upper 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. This happens for the transition from the highest possible energy level (approaching infinity, $n\_2\; =\; \backslash infty$) to the lowest level of the series ($n\_1\; =\; 3$).

Substituting these values into the Rydberg formula:

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

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

Now, we can calculate the shortest wavelength:

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

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

$$\backslash lambda\_\{min\}\; =\; \backslash frac\{9\}\{1.097\; \backslash times\; 10^7\; \backslash text\{\; m\}^\{-1\}\}\; \backslash approx\; 8.204\; \backslash times\; 10^\{-7\}\; \backslash text\{\; m\}$$

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

$$\backslash lambda\_\{min\}\; \backslash approx\; 8.204\; \backslash times\; 10^\{-7\}\; \backslash text\{\; m\}\; \backslash times\; \backslash frac\{10^9\; \backslash text\{\; nm\}\}\{1\; \backslash text\{\; m\}\}\; \backslash approx\; 820.4\; \backslash text\{\; nm\}$$

The shortest wavelength present in the Paschen series of spectral lines is approximately 820.4 nm.