Answer

(a) The sizes of the atoms taken in Thomson’s model and Rutherford’s model have the same order of magnitude.

(b) In the ground state of Thomson’s model, the electrons are in stable equilibrium. However, in Rutherford’s model, the electrons always experience a net force.

(c) A classical atom based on Rutherford’s model is doomed to collapse.

(d) An atom has a nearly continuous mass distribution in Thomson’s model, but has a highly non-uniform mass distribution in Rutherford’s model.

(e) The positively charged part of the atom possesses most of the mass in both the models.

Answer

In the alpha-particle scattering experiment, if a thin sheet of solid hydrogen is used in place of a gold foil, then the scattering angle would not be large enough. This is because the mass of hydrogen (1.67 × 10⁻²⁷ kg) is less than the mass of incident α−particles (6.64 × 10⁻²⁷ kg). Thus, the mass of the scattering particle is more than the target nucleus (hydrogen). As a result, the α−particles would not bounce back if solid hydrogen is used in the α-particle scattering experiment.

Answer

Rydberg’s formula is given as:

Where,

h = Planck’s constant = 6.6 × 10⁻³⁴ Js

c = Speed of light = 3 × 10⁸ m/s

(n₁ and n₂ are integers)

The shortest wavelength present in the Paschen series of the spectral lines is given for values n₁ = 3 and n₂ = ∞.

Answer

Separation of two energy levels in an atom,

E = 2.3 eV

= 2.3 × 1.6 × 10⁻¹⁹

= 3.68 × 10⁻¹⁹ J

Let ν be the frequency of radiation emitted when the atom transits from the upper level to the lower level.

We have the relation for energy as:

E = hv

Where,

h = Planck’s constant

Hence, the frequency of the radiation is 5.6 × 10¹⁴ Hz.

Answer

Ground state energy of hydrogen atom, E = − 13.6 eV

This is the total energy of a hydrogen atom. Kinetic energy is equal to the negative of the total energy.

Kinetic energy = − E = − (− 13.6) = 13.6 eV

Potential energy is equal to the negative of two times of kinetic energy.

Potential energy = − 2 × (13.6) = − 27 .2 eV

Answer

For ground level, n₁ = 1

Let E₁ be the energy of this level. It is known that E₁ is related with n₁ as:

The atom is excited to a higher level, n₂ = 4.

Let E₂ be the energy of this level.

The amount of energy absorbed by the photon is given as:

E = E₂ − E₁

For a photon of wavelengthλ, the expression of energy is written as:

Where,

h = Planck’s constant = 6.6 × 10⁻³⁴ Js

c = Speed of light = 3 × 10⁸ m/s

And, frequency of a photon is given by the relation,

Hence, the wavelength of the photon is 97 nm while the frequency is 3.1 × 10¹⁵ Hz.

Answer

(a) Let ν₁ be the orbital speed of the electron in a hydrogen atom in the ground state level, n₁ = 1. For charge (e) of an electron, ν₁ is given by the relation,

Where,

e = 1.6 × 10⁻¹⁹ C

∈₀ = Permittivity of free space = 8.85 × 10⁻¹² N⁻¹ C² m⁻²

h = Planck’s constant = 6.62 × 10⁻³⁴ Js

For level n₂ = 2, we can write the relation for the corresponding orbital speed as:

And, for n₃ = 3, we can write the relation for the corresponding orbital speed as:

Hence, the speed of the electron in a hydrogen atom in n = 1, n=2, and n=3 is 2.18 × 10⁶ m/s, 1.09 × 10⁶ m/s, 7.27 × 10⁵ m/s respectively.

(b) Let T₁ be the orbital period of the electron when it is in level n₁ = 1.

Orbital period is related to orbital speed as:

Where,

r₁ = Radius of the orbit

h = Planck’s constant = 6.62 × 10⁻³⁴ Js

e = Charge on an electron = 1.6 × 10⁻¹⁹ C

∈₀ = Permittivity of free space = 8.85 × 10⁻¹² N⁻¹ C² m⁻²

m = Mass of an electron = 9.1 × 10⁻³¹ kg

For level n₂ = 2, we can write the period as:

Where,

r₂ = Radius of the electron in n₂ = 2

And, for level n₃ = 3, we can write the period as:

Where,

r₃ = Radius of the electron in n₃ = 3

Hence, the orbital period in each of these levels is 1.52 × 10⁻¹⁶ s, 1.22 × 10⁻¹⁵ s, and 4.12 × 10⁻¹⁵ s respectively.

Answer

The radius of the innermost orbit of a hydrogen atom, r₁ = 5.3 × 10⁻¹¹ m.

Let r₂ be the radius of the orbit at n = 2. It is related to the radius of the innermost orbit as:

For n = 3, we can write the corresponding electron radius as:

Hence, the radii of an electron for n = 2 and n = 3 orbits are 2.12 × 10⁻¹⁰ m and 4.77 × 10⁻¹⁰ m respectively.

Answer

It is given that the energy of the electron beam used to bombard gaseous hydrogen at room temperature is 12.5 eV. Also, the energy of the gaseous hydrogen in its ground state at room temperature is −13.6 eV.

When gaseous hydrogen is bombarded with an electron beam, the energy of the gaseous hydrogen becomes −13.6 + 12.5 eV i.e., −1.1 eV.

Orbital energy is related to orbit level (n) as:

For n = 3, 

This energy is approximately equal to the energy of gaseous hydrogen. It can be concluded that the electron has jumped from n = 1 to n = 3 level.

During its de-excitation, the electrons can jump from n = 3 to n = 1 directly, which forms a line of the Lyman series of the hydrogen spectrum.

We have the relation for wave number for Lyman series as:

Where,

R_y = Rydberg constant = 1.097 × 10⁷ m⁻¹

λ= Wavelength of radiation emitted by the transition of the electron

For n = 3, we can obtain λas:

If the electron jumps from n = 2 to n = 1, then the wavelength of the radiation is given as:

If the transition takes place from n = 3 to n = 2, then the wavelength of the radiation is given as:

This radiation corresponds to the Balmer series of the hydrogen spectrum.

Hence, in Lyman series, two wavelengths i.e., 102.5 nm and 121.5 nm are emitted. And in the Balmer series, one wavelength i.e., 656.33 nm is emitted.

Answer

Radius of the orbit of the Earth around the Sun, r = 1.5 × 10¹¹ m

Orbital speed of the Earth, ν = 3 × 10⁴ m/s

Mass of the Earth, m = 6.0 × 10²⁴ kg

According to Bohr’s model, angular momentum is quantized and given as:

Where,

h = Planck’s constant = 6.62 × 10⁻³⁴ Js

n = Quantum number

Hence, the quanta number that characterizes the Earth’ revolution is 2.6 × 10⁷⁴.

Answer

(a) about the same

The average angle of deflection of α­-particles by a thin gold foil predicted by Thomson’s model is about the same size as predicted by Rutherford’s model. This is because the average angle was taken in both models.

(b) much less

The probability of scattering of α-particles at angles greater than 90° predicted by Thomson’s model is much less than that predicted by Rutherford’s model.

(c) Scattering is mainly due to single collisions. The chances of a single collision increases linearly with the number of target atoms. Since the number of target atoms increase with an increase in thickness, the collision probability depends linearly on the thickness of the target.

(d) Thomson’s model

It is wrong to ignore multiple scattering in Thomson’s model for the calculation of average angle of scattering of α­−particles by a thin foil. This is because a single collision causes very little deflection in this model. Hence, the observed average scattering angle can be explained only by considering multiple scattering.

Answer

Radius of the first Bohr orbit is given by the relation,

Where,

∈₀ = Permittivity of free space

h = Planck’s constant = 6.63 × 10⁻³⁴ Js

m_e = Mass of an electron = 9.1 × 10⁻³¹ kg

e = Charge of an electron = 1.9 × 10⁻¹⁹ C

m_p = Mass of a proton = 1.67 × 10⁻²⁷ kg

r = Distance between the electron and the proton

Coulomb attraction between an electron and a proton is given as: 

Gravitational force of attraction between an electron and a proton is given as:

Where,

G = Gravitational constant = 6.67 × 10⁻¹¹ N m²/kg²

If the electrostatic (Coulomb) force and the gravitational force between an electron and a proton are equal, then we can write:

∴F_G = F_C

Putting the value of equation (4) in equation (1), we get:

It is known that the universe is 156 billion light years wide or 1.5 × 10²⁷ m wide. Hence, we can conclude that the radius of the first Bohr orbit is much greater than the estimated size of the whole universe.

Answer

It is given that a hydrogen atom de-excites from an upper level (n) to a lower level (n−1).

We have the relation for energy (E₁) of radiation at level n as:

Now, the relation for energy (E₂) of radiation at level (n − 1) is givenas: 

Energy (E) released as a result of de-excitation:

E = E₂−E₁

hν = E₂ − E₁ … (iii)

Where,

ν = Frequency of radiation emitted

Putting values from equations (i) and (ii) in equation (iii), we get:

For large n, we can write

Classical relation of frequency of revolution of an electron is given as: 

Where,

Velocity of the electron in the n^th orbit is given as:

v =

And, radius of the n^th orbit is given as:

r = 

Putting the values of equations (vi) and (vii) in equation (v), we get:

Hence, the frequency of radiation emitted by the hydrogen atom is equal to its classical orbital frequency.

Answer

(a) Charge on an electron, e = 1.6 × 10⁻¹⁹ C

Mass of an electron, m_e = 9.1 × 10⁻³¹ kg

Speed of light, c = 3 ×10⁸ m/s

Let us take a quantity involving the given quantities as 

Where,

∈₀ = Permittivity of free space

And, 

The numerical value of the taken quantity will be:

Hence, the numerical value of the taken quantity is much smaller than the typical size of an atom.

(b) Charge on an electron, e = 1.6 × 10⁻¹⁹ C

Mass of an electron, m_e = 9.1 × 10⁻³¹ kg

Planck’s constant, h = 6.63 ×10⁻³⁴ Js

Let us take a quantity involving the given quantities as 

Where,

∈₀ = Permittivity of free space

And, 

The numerical value of the taken quantity will be:

Hence, the value of the quantity taken is of the order of the atomic size.

Answer

(a) Total energy of the electron, E = −3.4 eV

Kinetic energy of the electron is equal to the negative of the total energy.

K = −E

= − (− 3.4) = +3.4 eV

Hence, the kinetic energy of the electron in the given state is +3.4 eV.

(b) Potential energy (U) of the electron is equal to the negative of twice of its kinetic energy.

U = −2 K

= − 2 × 3.4 = − 6.8 eV

Hence, the potential energy of the electron in the given state is − 6.8 eV.

(c) The potential energy of a system depends on the reference point taken. Here, the potential energy of the reference point is taken as zero. If the reference point is changed, then the value of the potential energy of the system also changes. Since total energy is the sum of kinetic and potential energies, total energy of the system will also change.

Answer

We never speak of quantization of orbits of planets around the Sun because the angular momentum associated with planetary motion is largely relative to the value of Planck’s constant (h). The angular momentum of the Earth in its orbit is of the order of 10⁷⁰h. This leads to a very high value of quantum levels n of the order of 10⁷⁰. For large values of n, successive energies and angular momenta are relatively very small. Hence, the quantum levels for planetary motion are considered continuous.

Answer

Mass of a negatively charged muon, 

According to Bohr’s model,

Bohr radius, 

And, energy of a ground state electronic hydrogen atom,

We have the value of the first Bohr orbit, 

Let r_μ be the radius of muonic hydrogen atom.

At equilibrium, we can write the relation as:

Hence, the value of the first Bohr radius of a muonic hydrogen atom is

2.56 × 10⁻¹³ m.

We have,

E_e= − 13.6 eV

Take the ratio of these energies as:

Hence, the ground state energy of a muonic hydrogen atom is −2.81 keV.