Atoms MCQ Quiz - Objective Question with Answer for Atoms - 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

 Statement I: Atoms are electrically neutral as they contain equal number of proton and neutron

Statement II: Atoms of each element are stable and emit their characteristic spectrum.

The correct answer is option 2, Explanation: 

Statement I is incorrect because atoms are electrically neutral due to the equal number of protons (positive charges) and electrons (negative charges).

Statement II is incorrect because not all atoms of each element are inherently stable; some can be radioactive and unstable.

However, they do emit characteristic spectra when energy is absorbed and re-emitted.

Hence, the correct option is 2, which states that Statement I is incorrect but Statement II is incorrect.

Calculation:

\(E_n =-\dfrac{13.6}{n^2}\ eV \)

\(\Rightarrow E_5=\dfrac{-13.6}{5^2}=\dfrac {-13.6}{25}=-0.54\ eV\)

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)

CONCEPT:

• Rutherford’s Model of the Atom:
  • He bombarded the beam of alpha particles on a very thin gold foil.
  • While bombarding he observes the number of scattering of particles –
    • Most of the particles passed either un-deviated or with very small deviation.
    • Some deviated by large angles
    • 1 out of 8000 was deflected by more than 90°
  • Conclusion to this deflection:
    • Most of the space of an atom is empty.
    • At the center of an atom having a tiny positively charged particle called a nucleus.
    • The center nucleus has all the mass of an atom.
    • The amount of positive charge at the nucleus is equal to the total amount of negative charges on all the electrons of the atom.
    • All the electrons revolve around the nucleus and coulomb force provide centripetal force

EXPLANATION:

This the nucleus of the atom is discovered by Rutherford atomic model. So option 3 is correct.

• Limitations of Rutherford’s Model of the Atom:
  • As Rutherford’s Model suggests electrons always revolve around the nucleus.
  • From the point of mechanics it is okay as Coulomb force provides the necessary centripetal force.
  • But Maxwell’s equation of electromagnetism show any accelerated electron must continuously emit electromagnetic radiation.
  • This revolving electron continuously emit radiation at all temperature.
  • This energy spend ultimately makes an electron to fall on the nucleus.
  • But in real case hydrogen is very stable nor emit any energy.

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.

The correct answer is The empty space present in atom.

CONCEPT:

Rutherford’s Model of the Atom: 

• While bombarding, he observes the number of scattering of particles:
• Most of the particles passed either un-deviated or with a very small deviation and very few deviated by large angles
  • 1 out of 8000 was deflected by more than 90°
• The conclusion to this deflection:
  • Most of the space of an atom is empty and at the center of an atom having a tiny positively charged particle called a nucleus.
  • The center nucleus has all the mass of an atom and all the electrons revolve around the nucleus and coulomb force provides the centripetal force

EXPLANATION:

• From the above, it is clear that when α-particles are passed through a thin metal foil, most of them go straight through the foil because of the empty space present in the atom. Therefore option 1 is correct.

Additional Information

• Limitations of Rutherford’s Model of the Atom:
  • As Rutherford’s Model suggests electrons always revolve around the nucleus.
  • From the point of mechanics, it is okay as Coulomb's force provides the necessary centripetal force.
  • But according to Maxwell’s equation of electromagnetism, this revolving electron continuously emits radiation at all temperatures.
  • This energy spends ultimately makes an electron fall on the nucleus. But in the real case hydrogen is very stable nor emit any energy.

The correct answer is Gamma Rays.

Key Points

• Gamma Rays are high-frequency electromagnetic radiation.
• Ernest Rutherford named them "gamma rays".
• These are not deflected by electric and magnetic fields.
• It shows that they have no charge.
• These are electromagnetic waves like X-rays.
• Its photons have shorter wavelengths than X-rays.
• It travels at the speed of light.​

Additional Information

• ​​Cathode Rays -
  • ​​Cathode Rays travel in a straight line and these rays are made up of small matter particles.
  • These rays are negative in nature. Therefore, they are also called negative particles.
  • These rays are affected by the magnetic field and these rays affect the photographic plate.
  • On passing them through gases, they ionize the gases.​
• Alpha Rays -
  • These are positively charged particles. Its symbol is 'α'.
  • It is a highly active and energetic helium atom containing two neutrons and two protons.
  • It can be released from the nucleus of a radioactive atom.​
• Beta Rays -
  • These are extremely energetic electrons that are released from the inner nucleus.
  • Their mass is negligible and the charge is negative.
  • It is denoted by the Greek-character beta (β).
  • These rays are used in medical applications, such as the treatment of eye diseases.

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:

• When the electrons in the atoms are in the state other than the ground state then this is called the atom in an excited state.
• The lifetime of atoms in an excited state is the time duration in which the electrons remain in their excited state.
  • The lifetime of atoms in an excited state is an average lifetime derived from the decay probability.
• Excited-state lifetimes are typically in few nanoseconds, The closest answer is 10^-8 seconds. So option 1 is correct.

Copper atomic number 29.

• Normally it has 29 (2, 8, 18, 1) electrons, 29 protons, and 35 neutrons.
• It belongs to Group 11.

Important Points

Oxygen atomic number 8.

• Normally it has 8 (2, 6) electrons, 8 protons, and 8 neutrons.
• It belongs to Group 16 (Chalcogens group).

Nitrogen atomic number 7.

• Normally it has 7 (2, 5) electrons, 7 protons, and 7 neutrons.
• It belongs to Group 15 (Pnictogen group).​

Additional Information

The hydrogen (H) atom does not have any neutrons in its tiny nucleus.

• It has only one electron (-ve charged) and one proton (+ve charged).
• The first isotope of Hydrogen called Protium has no neutron in its atom.
• Neutrons are the particles in an atom that has a neutral charge.

The correct answer is option 4) i.e. cutoff wavelength

CONCEPT:

• X-ray spectrum: The X-ray spectrum is defined as the energy distribution of X-ray radiation.
  • X-rays are produced when an electrons incident on the target material inside an X-ray tube emits electromagnetic radiation.
  • There are two types of X-rays produced - characteristic lines and continuous spectrum.
  • Characteristic lines are produced due to the transition of an electron within the atom of the target material.
  • A continuous spectrum is produced due to electrons colliding with the target and decelerating. Therefore, the continuous spectrum has a minimum wavelength.
  • X- rays are emitted when the kinetic energy of the electrons is completely converted into photon energy.

KE of electron = Photon energy

eV = \(\frac{hc}{λ}\)

Where λ is the minimum wavelength.

Minimum wavelength indicates the most energetic photon emitted due to a highly energetic electron, that collides with the target and decelerates to stop immediately after a collision.

EXPLANATION:

• When the target material has changed

1. the intensity of the X-ray spectra varies.
2. The wavelength of characteristic lines varies.
3. The minimum wavelength or the cutoff wavelength remains unchanged.

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.