De Broglie’s Explanation of Bohr’s Second Postulate of Quantisation MCQ Quiz - Objective Question with Answer for De Broglie’s Explanation of Bohr’s Second Postulate of Quantisation - Download Free PDF

Correct option is: (3) r ∝ n^2/3, v ∝ n^1/3

Given, force is constant

F = mv² / r

⇒ v² / r = constant

⇒ r ∝ v²      ...(1)

L = mvr = nh / 2π      ...(2)

On solving equation (1) and equation (2):

v ∝ n^1/3 and r ∝ n^2/3

Concept:

De-Broglie waves explain the nature of the wave related to the particle. For particles with mass like electrons, protons, etc., to calculate wavelength, de Broglie wavelength formula is used.

At non-relativistic speeds( less than the speed of light), the de-Broglie wavelength is calculated using the following formula:

λ = \(\rm\frac{h}{mv}\)

where "h" is Planck's constant, m and v are the mass and velocity of the particle respectively.

 The S.I. unit of the de Broglie wavelength is in meters. 

Calculation:

De-Broglie wavelength

λ = \(\rm\frac{h}{mv}\)

= \(\frac{h}{\sqrt{2m(KE)}}\)

\(=\frac{h}{\sqrt{2m\left(\frac{3}{2}kT\right)}}\)

\(\lambda - \frac{h}{\sqrt{3mkT}}\)

Explanation:

For a photon,

(i) Energy E = h v ⇒  (statement A is correct)

(ii) All photons travel with speed of light (c in free space)  statement B is correct

(iii) Momentum of a photon. p = hv / c, Statement C is correct.

(iv) In a photon-electron collision, total energy and total momentum are conserved, statement D is also correct.

(v) Photons are massless and do not carry any charge,statement E is incorrect.

∴ The correct option is 2)

Calculation:

De-Broglie wavelength and energy relation of a free particle 

λ² = h √ 2 m E 

λ² = h²/2 m E

1/λ² = 2 mE/h2

∴ The correct option is 4)

Concept Used:

The de Broglie wavelength (λ) of a particle is given by:

λ = h / p

where h is Planck’s constant and p is the momentum of the particle.

Momentum (p) is related to kinetic energy (K) as:

p = √(2mK)

where m is the mass and K is the kinetic energy of the particle.

Calculation:

Given:

Mass of particle A: M_A

Mass of particle B: M_B

M_A = 2M_B

Kinetic energy of particle A: K_A

Kinetic energy of particle B: K_B = K_A / 8

Now, the de Broglie wavelength for particle A:

λ_A = h / √(2 M_A K_A)

And for particle B:

λ_B = h / √(2 M_B K_B)

Taking the ratio:

λ_A / λ_B = √(M_B K_B / M_A K_A)

Substituting values:

λ_A / λ_B = √(M_B × (K_A / 8) / (2M_B K_A))

λ_A / λ_B = √(1 / 16)

λ_A / λ_B = 1 / 4

∴ The correct answer is 1:4.

CONCEPT:

• de Broglie wavelength of electrons: Louis de Broglie theorized that not only light possesses both wave and particle properties, but rather particles with mass - such as electrons - do as well.
  • The wavelength of material waves is also known as the de Broglie wavelength.
• de Broglie wavelength of electrons can be calculated from Planks constant h divided by the momentum of the particle.

λ = h/p

where λ is de Broglie wavelength, h is Plank's const, and p is the momentum.

In terms of Energy, de Broglie wavelength of electrons:

\(λ=\frac{h}{\sqrt{2mE}}\)

where λ is de Broglie wavelength, h is Plank's const, m is the mass of an electron, and E is the energy of the electron.

• Electric Potential energy: If a Charge q is placed in the electric potential V, the electric potential energy

U = E = qV

where U or E is the electric potential energy, q is the charge, and V is the electric potential.

CALCULATION:

\(λ=\frac{h}{\sqrt{2mE}}\)

\(λ=\frac{h}{\sqrt{2mqV}}\)

\(λ_1=\frac{h}{\sqrt{2m_pq_pV}}\)

m_p and q_p is the mass and charge of the proton.

\(λ_2=\frac{h}{\sqrt{2m_α q_α V}}\)

mα and qα is the mass and charge of the alpha particle.

\(\frac{λ_1}{\lambda_2}=\sqrt\frac{m_\alpha q_\alpha}{m_pq_p}\)

4m_p =  mα  and 2q_p = qα

\(\frac{λ_1}{\lambda_2}=\sqrt\frac{4m_p 2q_p}{m_pq_p}\)

\(\frac{λ_1}{\lambda_2}=\sqrt8\)

\(\frac{λ_1}{\lambda_2}=2\sqrt2\)

So, the correct answer is option 1.

Concept:

De Broglie proposed that the wavelength λ associated with a particle of momentum p is given by:

\(\lambda = \frac{h}{p} = \frac{h}{{mv}}\)

where m is the mass of the particle and v its speed. Planck’s Constant, h = 6.623 × 10^-34 Js,

Above equation is known as the de Broglie relation and the wavelength λ of the matter-wave is called de Broglie wavelength.

Thus, the significance of de Broglie equation lies in the fact that it relates the particle character to the wave character of matter.

Important Points

According to Planck’s quantum theory, the energy of a photon is described as E = hν

According to Einstein’s mass energy relation, E = mc²

Frequency ν can be expressed in terms of wavelength λ as, ν = c/λ

hν = mc² ⇒ hν/c = mc ⇒ λ = h/mc, (This equation is applicable for a photon)

Calculation:

Given:

m = 9.11 × 10-31 kg, v = 6 × 107 m/s

\(\lambda = \frac{h}{{mv}} = \frac{{6.623 \times {{10}^{ - 34}}}}{{9.11 \times {{10}^{ - 31}} \times 6 \times {{10}^7}}} = 1.21 \times {10^{ - 11}}m = 0.121 \) Å

Concept:

According to the De-Broglie hypothesis, particles behave as waves & these are called matter waves. The wavelength of the particle is called De-Broglie wavelength and it is given as

\(⇒\lambda = \frac{h}{\sqrt{2mKE}}\)

Where m = mass of the charged particle and KE = Kinetic energy

Calculation:

• The de-Broglie wavelength is given by:

\(⇒\lambda = \frac{h}{\sqrt{2mKE}}\)

By squaring both sides of the above equation, we get:

\(⇒ KE = \frac{h^{2}}{2m\lambda^{2}}\)

• The kinetic energy of the electron is given by

\(⇒ KE_{e} = \frac{h^{2}}{2m_{e} \lambda^{2}}\)     -----(1)

• The kinetic energy of the proton is given by

\(⇒ KE_{p} = \frac{h^{2}}{2m_{p} \lambda^{2}}\)     -----(2) 

• By dividing equation 1 and equation 2:

\( ⇒ \frac{KE_{e}}{KE_{p}}=\frac{m_{p}}{m_{e}}\)

As we know,

⇒ m_P > m_e

So, KE_e > KE_p

Hence, option 3 is the answer

CONCEPT:

Matter waves (de-Broglie Waves): According to de-Broglie a moving material particle sometimes acts as a wave and sometimes as a particle.

A wave is associated with a moving material particle which controls the particle in every respect.

EXPLANATION:

de-Broglie wavelength: According to de-Broglie theory, the wavelength of a de-Broglie wave is given by:

\(\lambda = \frac{h}{p} = \frac{h}{{mv}} = \frac{h}{{\sqrt {2mE} }}\)

Thus,

\(\Rightarrow \lambda \propto \frac{1}{p} \propto \frac{1}{v} \propto \frac{1}{{\sqrt E }}\)

Where h = Plank's constant, m = Mass of the particle, v = Speed of the particle, E = Energy of the particle

From the equation above, it is clear that the wavelength of the electron is inversely proportional to the kinetic energy. Therefore, when the kinetic energy of an electron is increased, the wavelength of the associated wave will decrease. Hence option 2 is correct.

The smallest wavelength whose measurement is possible is that of γ -rays.

The wavelength of matter waves associated with the microscopic particles like electron, proton, neutron, α -particle, etc. is of the order of 10 ^-10 m.

CONCEPT:

• Louis de Broglie in his theory on wave nature of matter proposed that:
  • All particles could be treated as matter waves with a wavelength $λ.$
• And their frequency is given by the following equation:

\(λ = \frac{h}{mv}\)

where λ is de Broglie wavelength, h is Planck's const, m is mass, v is the velocity.

\(λ = \frac{h}{\sqrt{2mE_k}}\)

where λ is de Broglie wavelength, h is Planck's const, m is mass, E_k is the Kinetic energy.

• Mass of Proton = 1.673 × 10^-27 Kg; 
• Mass of nutron = 1.675 × 10-27 Kg;
• Mass of α Particle = 4 × 1.673 × 10-27 Kg
• Mass of electron = 9.1 × 10-31 Kg

EXPLANATION: 

• Given that the Kinetic energy of all the particles is the same. So

\(λ = \frac{h}{\sqrt{2mE_k}} α \frac{1}{\sqrt m}\)

• The above equation says the greater the mass the lesser the wavelength.
• Mass of α particle is greatest in the given option, So it will have the least wavelength.
• So the correct answer is option 2.

Concept:

According to the De-Broglie hypothesis, particles behave as waves & these are called matter waves. The wavelength of the particle is called De-Broglie wavelength and it is given as

\(⇒\lambda = \frac{h}{\sqrt{2mKE}}\)

Where m = mass of the charged particle and KE = Kinetic energy

Calculation:

A deuteron and α-particle have the same kinetic energy then KE₁ = KE₂

\(⇒\lambda = \frac{h}{\sqrt{2mKE}}\)

\(\frac{{{\lambda _d}}}{{{\lambda _\alpha }}} = \frac{{\sqrt {{m_\alpha }} }}{{\sqrt {{m_d}} }}\)

= \(\frac{{\sqrt {4{m_p}} }}{{\sqrt {2{m_p}} }} = \sqrt 2\)

Where; m_p is the mass of a proton.

Note:

1). Mass of deuteron is 2 times of proton mass and the Charge of the deuteron is equal to the charge of the proton.

2). Mass of α-particle is 4 times of proton mass and the Charge of the α-particle is equal to the 2 times of charge of the proton.

CONCEPT:

• de Broglie wavelength of electrons: Louis de Broglie theorized that not only light possesses both wave and particle properties, but rather particles with mass - such as electrons - do as well.
  • The wavelength of material waves is also known as the de Broglie wavelength.
• de Broglie wavelength (λ) of electrons can be calculated from Planks constant h divided by the momentum of the particle.

λ = h/p

where h is Plank's const, λ is de Broglie wavelength, and p is the momentum.

• In terms of Kinetic energy, de Broglie wavelength of electrons:

\(λ=\frac{h}{\sqrt{2mE}}\)

where λ is de Broglie wavelength, m is the mass of an electron, h is Plank's const, and E is the energy of the electron.

CALCULATION:

Given that: 

New Kinetic energy (E₂) = 16 times the given kinetic energy = 16 E₁

\(λ_1=\frac{h}{\sqrt{2mE_1}}\)

\(λ_1=\frac{h}{\sqrt{2mE_2}}=\frac{h}{\sqrt{2m\;\times\; 16E_1}} =\frac{1}{4}\lambda_1\)

The correct answer is option 1) i.e. λγ  > λα > λβ 

CONCEPT:

• Spectral lines are produced when electrons of an excited atom move between the orbital energy levels and return towards the ground state.
  • Each orbit of an atom is denoted by n = 1,2,3,4... and so on. Where n =1 is the ground state and n > 1 represents orbits of higher energy levels.
• Rydberg's formula gives a relationship between the energy level difference in various levels of Bohr's model of an atom and the wavelength of absorbed or emitted photons during the excitation of electrons.
  • Rydberg formula: \(\frac{1}{\lambda} = RZ^2 [{ \frac{1}{n_i^2} - \frac{1}{n_f^2}}]\)

Where R is the Rydberg constant, Z is the atomic number, n_i is the lower energy level and n_f is the higher energy level.

CALCULATION:

Given that:

Three spectral lines kα, kβ, and kγ are produced.

The possible electron movement is as follows:

Using Rydberg's formula: \(\frac{1}{\lambda} = RZ^2 [{ \frac{1}{n_i^2} - \frac{1}{n_f^2}}]\)

For k_α: \(\frac{1}{\lambda_\alpha} = RZ^2 [{ \frac{1}{1^2} - \frac{1}{2^2}}]\)\(= \)  0.75 RZ²      ----(1)

For k_β: \(\frac{1}{\lambda_\beta} = RZ^2 [{ \frac{1}{1^2} - \frac{1}{3^2}}]\)\(= \) 0.889 RZ²      ----(2)

For k_γ: \(\frac{1}{\lambda_\gamma} = RZ^2 [{ \frac{1}{2^2} - \frac{1}{3^2}}]\) \(=\)  0.138 RZ²      ----(3)

From (1), (2), and (3) we get, \(\frac{1}{\lambda_\beta}>\frac{1}{\lambda_\alpha} >\frac{1}{\lambda_\gamma}\)

Therefore, λγ  > λα > λβ 

Key Points

de Broglie wavelength:

• Louis de Broglie theorized that not only light possesses both wave and particle properties, but rather particles with mass - such as electrons - do as well.
• The wavelength of material waves is also known as the de Broglie wavelength.
• de Broglie wavelength can be calculated from Planks constant h divided by the momentum of the particle.

\(\lambda = \frac {h}{mv}\)

where λ is de Broglie wavelength, h is Plank's constant, and mv is the momentum.

• If different particles have the same velocity, then the wavelength is inversely proportional to the mass of that particle.

Calculation:

Given particles are proton, electron, neutron and alpha particle.

Now

The mass of proton = 1.67 × 10-27 kg,

The mass of neutron = 1.67 × 10-27 kg,

The mass of electron = 9.11 × 10-31 kg,

The mass of alpha particle = 6.65 × 10-27 kg;

Among these four particles, the Lightest particle is an electron;

∴ Hence an electron will have the longest wavelength.

CONCEPT:

De Broglie wavelength: It is a wavelength expressed in all the objects in quantum mechanics that determines the probability density of finding the object at a given point of the configuration space.

•  The de Broglie wavelength of a particle is inversely proportional to its momentum.
• Kinetic energy is

K = ½ m v², Where K = kinetic energy, m = mass of particle, v = velocity of particle.

• Momentum of particle

P = m v, Where p = momentum of particle,

• The de Broglie wavelength is given by

\({\rm{\lambda \;}} = \frac{{\rm{h}}}{{{\rm{m\;v}}}}\) Where λ = de Broglie wavelength, h = plank constant.

EXPLANATION:

Kinetic energy is given by:

K = ½ m v², Where K = kinetic energy, m = mass of particle, v = velocity of particle

\({\rm{v}} = \sqrt {{\rm{\;}}\frac{{2{\rm{\;K}}}}{{\rm{m}}}} \)

∴ \({\rm{p}} = {\rm{m\;v}} = {\rm{m}}\sqrt {{\rm{\;}}\frac{{2{\rm{\;K}}}}{{\rm{m}}}} \) And K = e V, Where V = voltage applied.

\({\rm{p}} = \sqrt {2{\rm{\;m\;e\;V}}} \), So the wavelength of the particle is

\({\rm{\lambda \;}} = \frac{{\rm{h}}}{{\sqrt {2{\rm{\;m\;e\;V}}} }}\), So the wavelength is inversely proportional to the square root of voltage. That is -

\({\rm{\lambda }} \propto \frac{1}{{\sqrt {\rm{V}} }}\)

So option 1 is correct.