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The nucleus of hydrogen contains only one proton. Its mass is 1.67 x 10^-27 kg. The mass of an alpha-particle is 6.64 x 10^-27 kg. When the alpha particles are incident on a thin sheet of solid hydrogen [in place of the gold foil], even in a head-on collision, the alpha-particle will not be scattered back. It is because the scattering particle is more massive than the target particle.

The first, second and third excited energies for hydrogen atom are
E’₁ = E₂ – E₁ = -3.4 – [-13.6] = 10.2 eV,
E’₂ = E₃ – E₁ = -1.51 – [-13.6] = 12.09 eV,
and E’₃ = E₄ – E₁ = -0.85 – [-13.6] = 12.75 eV
It follows that when a 12.5 eV electron beam is used to bombard gaseous hydrogen at room temperature, the H – atoms at the most will be raised to n = 3 level. The H – atoms may come to ground state directly or via n = 2 level. Therefore, the wavelengths emitted will lie in Balmer series [for de-excitation from n = 3 level to n = 2 and 1 level] and Lyman series [for de­excitation from n = 2 level to n = 1 level].

Thomson’s model predicts the backward scattering of a -particles much less than that predicted by Rutherford’s model. The reason is that in Thomson’s model, positive charge is uniformly distributed over whole of the nucleus instead of being concentrated in a small part.

On the basis of Thomson’s atom model, the scattering angle of α-particles due to single collision will be very small. It is because, mass of atom is distributed all over it. The observed average scattering angle can be explained only by considering multiple scattering. Therefore, it is wrong to ignore multiple scattering in Thomson’ atom model.

Since the mass of the atom is concentrated in the form of nucleus in Rutherford’s atom model, most of the scattering comes through a single collision and multiple scattering can be ignored.

Thomson’s atom model cannot explain the scattering of α -particles through large angles and particularly through 180°. It is because, the light particles like electrons cannot deflect the massive a -particles through 180°.

The first model of the atom with a substructure was put forward in 1898 by Sir J.J. Thomson (1856-1940), a British physicist. According to this model, an atom consists of a sphere with a uniform distribution of positive charge and electrons embedded in it such that the atom is electrically neutral and stable.

Drawbacks : This model, known as the plumpudding model, failed to account for the observed scattering of α-particles and spectra of various elements.

[Note : It can be shown that the Thomson atom cannot be stable.]

Correct option is (D) \(\cfrac{36}{5R}\)

1. Magnesium : IIA

2. Carbon : IVA

3. Sulphur : VIA

4. Neon : Zero

\(\bar v=\frac{1}{λ}= R_H\Big[\frac{1}{n^2_1}-\frac{1}{n_2^2}\Big]\)

Longest wavelength n₁ = 2 and n₂ = 3

\(=\frac{1}{λ}= R_H\Big[\frac{1}{n^2_1}-\frac{1}{n_2^2}\Big]\) \(R_H\Big[\frac{1}{2^2}-\frac{1}{3^2}\Big]\)

λ = \(\frac{1}{R_H\Big[\frac{1}{2^2}-\frac{1}{3^3}\Big]}\) = 6563 A⁰

Shortest Wavelength n₁ = 2 and n₂ = α

\(=\frac{1}{λ}= R_H\Big[\frac{1}{n^2_1}-\frac{1}{n_2^2}\Big]\) = \(R_H\Big[\frac{1}{2^2}-\frac{1}{\alpha^2}\Big]\)

λ =\(\frac{1}{R_H\Big[\frac{1}{2^2}-\frac{1}{\alpha^2}\Big]}\) = \(\frac{1}{1.097\times10^7\Big(\frac{1}{4}\Big)}\)

= 3646 A⁰

The potential energy (E_p) of the electron in an orbit is equal to twice its total
energy (E)
= -13.6 x 2 = -27.2eV
The kinetic energy (E_k) of the electron in an orbit is equal to negative of its total energy (E)
∴ E_k – E
= – (-13.6) = 13.6 eV.

Correct option : (3) 130 eV

Explanation : 

The electron undergoes an elastic collision with the concerned atom. Since 130 eV cannot raise the electron from ground state to the first excited state, the electron does not accept any energy from the electron. Hence the electron will emerge after interaction with the atom with 130 eV.