The Bohr Model of Hydrogen MCQ Quiz - Objective Question with Answer for The Bohr Model of Hydrogen - Download Free PDF

Concept:

The potential energy of an electron in the electric field of the nucleus (Coulombic attraction) in a hydrogen atom is given by the electrostatic potential energy formula.

Let:

'e' be the charge of the electron

'r' be the distance of the electron from the nucleus

 be the permittivity of free space

The electrostatic potential energy is given by:

The negative sign indicates that the force is attractive in nature between the positively charged nucleus and the negatively charged electron.

Correct Answer:

Calculation:

Second excited state → first excited state

n = 3 → n = 2

⇒ hc / λ₀ = 13.6 × (1/2² − 1/3²)  ...(i)

Third excited state → second orbit

n = 4 → n = 2

⇒ hc / (20λ₀/x) = 13.6 × (1/2² − 1/4²)  ...(ii)

Divide (ii) by (i):

x / 20 = (1/2² − 1/4²) / (1/2² − 1/3²)

⇒ x = 27

Calculation:

The centripetal force of the rotating electron is given by 

But, according to Bohr,

i.e. 

⇒ 

Concept:

Energy of a Photon: The energy of the photon, Ephoton, is given by:

E_photon = Work function + Kinetic energy of the ejected electron

The work function is the minimum energy required to eject an electron from the metal surface, and the kinetic energy of the ejected electron is determined by the stopping voltage.

The energy levels of the hydrogen atom are quantized, and the energy of the electron in the nth energy level is given by:

E_n = -13.6 / n² eV

Calculation:

Given,

Work function of the metallic electrode = 0.5 eV

Stopping voltage = 0.47 V

The total energy of the emitted photon is:

E_photon = Work function + Kinetic energy of ejected electron

E_photon = 0.5 eV + 0.47 eV = 0.97 eV

The energy difference between the two states (n and the second excited state n = 3) is:

ΔE = E_n - E₃ = -13.6 * (1/n² - 1/3²) eV

Substituting the photon energy:

0.97 = -13.6 * (1/n² - 1/9)

⇒ -0.97 = 13.6 * (1/n² - 1/9)

⇒ -0.07132 = (1/n² - 1/9)

⇒1/n² = -0.07132 + 0.1111 = 0.03978

⇒ n² = 1 / 0.03978 ≈ 25.14

⇒ n ≈ √25.14 ≈ 5

∴ The quantum number of the state 'n' is approximately 5.
The correct option is 1) 5.

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.

CONCEPT:

• Bohr’s Model: Niels Bohr by making 4 postulates solved the puzzle of hydrogen spectra.
  • The mass of the nucleus is very large compared to that of the electrons and almost the entire mass of the atom is concentrated in the nucleus and hence assumed to be infinite.
  • The electrons revolve around the nucleus in circular orbits.
  • The mass of the electron remains constant.
  • The radius around the orbit has some special values of the radius. In these stationary orbits, they do not radiate energy as expected from Maxwell’s laws.
  • The energy of each stationary orbits are fixed, electrons can jump from a higher orbit to a lower orbit by emitting a photon of radiation. Where Higher energy orbit – Lower energy orbit = (h c)/λ  
    • An electron can also jump from lower to higher energy by absorbing energy.
  • In stationary orbits, the angular momentum L of an electron is an integral multiple of (h/2π)
    • L = n × (h/2π)

EXPLANATION:

• From the above, it is clear that all the above conditions are true. Hence option 4 is correct.

CONCEPT:

• Magnetic dipole moment. It is a magnetic property of electric current loops or magnets.
• The quantity of magnetic dipole moment is equal to the amount of current flowing in the loop multiplied by the area that the loop encompasses.

Magnetic dipole moment (μ) = IA

where I is current and A is the area.

• Angular momentum: the quantity of a rotating body, which is the product of its moment of inertia and its angular velocity.

Angular momentum (L) = m v r

where m is the mass of the body, v is velocity and r is the distance from the rotating point.

From Bohr's Model of Atom:A

​angular momentum (L) = mvr = nh/2π

v = nh/2πmr

Time period T = 2πr/v

current I = q/T = e/T = e/(2πr/v) = 

where n is the orbit number, h is plank constant, m is the mass of electron, r is the radius of orbit, v is the velocity of the electron, e is the charge on one electron.

CALCULATION:

Magnetic dipole moment (μ) = IA

μ =  = 

angular momentum (L) = nh/2π

So the correct answer is option 1.

Ans: Option: 1

CONCEPT: 

• Angular Momentum: The moment of momentum is called angular momentum.

  • It is the property of any rotating object given by moment of inertia times angular velocity.

• Bohr's postulate 2: In a hydrogen atom, the electron can revolve around the nucleus, without radiating energy, only in those orbits for which the angular momentum of the electron is equal to an integral multiple  of where h is Planck's constant.

Angular momentum (J) = m v r = 

Where m is the mass of the electron, v is the velocity of the electron and r is the radius of the orbit

EXPLANATION:

Angular momentum (J) = m v r = 

by solving we get,  

CONCEPT:

• The radius of a hydrogen-like atom is given by:

r = a0n2/Z

where r is the radius of atom, Z is the atomic number of the atom, n is the orbit number, and a0 is the radius of the 1st orbit of hydrogen.

CALCULATION:

Given -  Z = 1, and n = 3

• The radius of the 1^st orbit of hydrogen atoms given by:

⇒ r₁ = a_o   

• The radius of the 3^rd orbits of hydrogen atoms given by:

⇒ r₃ = ao32/Z = a_o9       

⇒ r₃ = 9r               [∵ r = a₀] 

Concept:

Bohr's Atomic Model –

• Bohr proposed a model for hydrogen atom which is also applicable for some lighter atoms in which a single electron revolves around a stationary nucleus of positive charge Ze (called hydrogen-like atom).

Bohr's model is based on the following postulates –

• He postulated that an electron in an atom can move around the nucleus in certain circular stable orbits without emitting radiations.
• Bohr found that the magnitude of the electron's angular momentum is quantized
• i.e. 
• Where n = 1, 2, 3, ..... each value of n corresponds to a permitted value of the orbit radius, r_n = Radius of nth orbit, v_n = corresponding speed.
• The radiation of energy occurs only when an electron jumps from one permitted orbit to another.

EXPLANATION:

From above it clear that the angular momentum of electron in n th orbit is given by . Thus option 3 is correct.

CONCEPT:

• Bohr model: In 1913, Niels Bohr gave the Bohr's atom model which retained essential features of Rutherford's model and at the same time took into account its drawbacks.
• The electrons revolve around the nucleus in circular orbits which is called the stationary orbit.
• According to the Bohr's atom model, the electrons of an atom revolve around the nucleus only in those orbits in which the angular momentum of the electron is an integral multiple of h /2π.​
• By absorbing energy the electrons are able to jump from lower energy(E_i) level to higher energy level (E_f) and vice versa.
• The energy of emitted radiation is given by 

• The energy of electrons in any orbit is given by:

Where n is principal quantum number and Z is the atomic number.

CALCULATION:

Given - Ground state energy = -13.6 eV, n =2 (Since first excited state) Z =2

• The energy of the n^th excited state can be calculated using the equation 

• The Energy of the first excited state will be 

CONCEPT:

Bohr's Atomic Model:

• Bohr proposed a model for hydrogen atom which is also applicable for some lighter atoms in which a single electron revolves around a stationary nucleus of positive charge Ze (called hydrogen-like atom).

Bohr's model is based on the following postulates:

• He postulated that an electron in an atom can move around the nucleus in certain circular stable orbits without emitting radiations.
• Bohr found that the magnitude of the electron's  angular momentum is quantized i.e.

Where n = 1, 2, 3, ..... each value of n corresponds to a permitted value of the orbit radius, rn = Radius of nth orbit, vn = corresponding speed and h = = Planck's constant

• The radiation of energy occurs only when an electron jumps from one permitted orbit to another.

​EXPLANATION:

• According to the Bohr model of the hydrogen atom, the radius of the n^th orbit is given as,

⇒ r_n = n² r₁     -----(1)

where n = number of orbit and r₁ = radius of the first orbit

• From equation 1 it is clear that the radius is directly proportional to the n².
• Hence, option 4 is correct.

CONCEPT:

The radius of a hydrogen-like atom is given by:

r = a₀n²/Z

where r is the radius of atom, Z is the atomic number of the atom, n is the orbit number, and a₀ is the radius of the 1st orbit of hydrogen.

CALCULATION:

Given that a₀ = 5.29 × 10-11 m; Z = 1, n = 2

r = a0n2/Z 

r = (5.29 × 10-11) ×2² /1

r = 21.16 × 10-11 m 

So the correct answer is option 3.

Explanation:

• The Bohr radius is the mean radius of the orbit of an electron around the nucleus of a hydrogen atom at its ground state (lowest energy level) 
• Bohr radius is symbolized by a
•  = 5.29 × 10^-11 m
• Where, ϵ0 = 8.854 × 10^-12 Fm is the vacuum permittivity of
• ̅h =  = 1.054 × 10^-34 Js is the reduced Plank's constant
• m_e = 9.109 × 10^-31 kg is the mass of electron
• e = 1.6 × 10^-19 C is the charge of electron
• c = 3 × 10⁸ m/s is the speed of light