Ans (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, ν1 = e²/n1 4π ε0 (h/2π) = e²/2 n1ε0h Where, e = 1.6 x 10⁻19 C ε0 = Permittivity of free space = 8.85 x 10⁻12 N-1 C2 m2 h = Planck’s cRead more
Ans (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,
ν1 = e²/n1 4π ε0 (h/2π) = e²/2 n1ε0h
Where, e = 1.6 x 10⁻19 C
ε0 = Permittivity of free space = 8.85 x 10⁻12 N-1 C2 m2
h = Planck’s constant = 6.62 x 10⁻³⁴ Js
Therefore, ν1 = (1.6 x 10⁻19 )2 /2 x (8.85 x 10⁻12) x (6.62 x 10⁻³⁴)
= 0.0218 x 10⁸ = 2.18 x 10⁶ m/s
For level n2= 2 ,we can write the relation for the corresponding orbital speed as:
ν2 = = e²/2 n2ε0h = (1.6 x 10⁻19 )2 /2 x 2 x(8.85 x 10⁻12) x (6.62 x 10⁻³⁴)
=1.09 x 10⁶ m/s
For level n3= 3 ,we can write the relation for the corresponding orbital speed as:
ν3 = = e²/2 n3ε0h = (1.6 x 10⁻19 )2 /2 x 3 x (8.85 x 10⁻12) x (6.62 x 10⁻³⁴)
=7.27 x 10⁵ m/s
Hence, the speed of the electron in a hydrogen atom in n = 1, n=2, and n=3 is 2.18 x 106 m/s, 1.09 x 106 m/s, 7.27 x 105 m/s respectively.
Ans (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:
T1 = 2π r/ν1
Where,r1 = Radius of the orbit =( n1)²h²ε0/πme²
h = Planck’s constant = 6.62 x 10⁻³⁴ Js and
e = Charge on an electron = 1.6 x 10⁻19 C.
ε0 = Permittivity of free space = 8.85 x 10⁻12 N⁻1C2 m⁻²
m = Mass of an electron = 9.1 x 10⁻31 kg
Therefore, T1 = 2π r1/ν1
= 2π x (1)² x (6.62 x 10⁻³⁴)² x (8.85 x 10⁻12)/[(2.18 x 10⁶) x π x ( 9.1 x 10⁻31) x (1.6 x 10⁻19)² ]
= 15.27 x 10⁻¹⁷ = 1.527 x 10⁻¹⁶ s
For level n2 = 2, we can write the period as:
T2 = 2π r2/ν2
Where, r2 = Radius of the electron in n2 = 2
= (n2)²h²ε0/πme²
Therefore,T2 =2π r2/ν2
=2π x (2)² x (6.62 x 10⁻³⁴)² x (8.85 x 10⁻12)/[(1.09 x 10⁶) x π x ( 9.1 x 10⁻31) x (1.6 x 10⁻19)² ]
=1.22 x 10⁻¹⁵ s
For level n3 = 3, we can write the period as:
T3 = 2π r3/ν3
Where, r3 = Radius of the electron in n3 = 3
= (n3)²h²ε0/πme²
Therefore,T3 =2π r3/ν3
=2π x (3)² x (6.62 x 10⁻³⁴)² x (8.85 x 10⁻12)/[(7.27 x 10⁵) x π x ( 9.1 x 10⁻31) x (1.6 x 10⁻19)² ]
=4.12 x 10⁻¹⁵ s
Hence, the orbital period in each of these levels is 1.527 x 10⁻¹⁶, 1.22 x 10⁻¹⁵ s, and 4.12 x 10⁻¹⁵ s respectively.
For ground level, n₁ = 1 Let E₁be the energy of this level. It is known that E₁ is related with m as: E₁ =-13.6/n²₁ eV = -13.6 /1² = -13.6 eV 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: Therefore E2 =-13.Read more
For ground level, n₁ = 1
Let E₁be the energy of this level. It is known that E₁ is related with m as:
E₁ =-13.6/n²₁ eV
= -13.6 /1² = -13.6 eV
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: Therefore E2 =-13.6/n²2 eV
=-13.6/4² = -13.6 /16 eV
The amount of energy absorbed by the photon is given as :
E = E2-E1
= -13.6 /16 -(-13.6 )
= 13.6 x 15/16 eV
= 13.6 x 15/16 x 1.6 x 10⁻¹⁹ = 2.04 x 10⁻¹⁸ J
For a photon of wavelength λ, the expression of energy is written as:
E = hc/λ
Where,
h = Planck’s constant = 6.6 x 10⁻34 Js and c = Speed of light = 3 x 10⁸ m/s.
Therefore , λ = hc/E = (6.6 x 10⁻34) x (3 x 10⁸ ) /(2.04 x 10⁻¹⁸ )
= 9.7 x 10⁻⁸ m = 97nm
And, frequency of a photon is given by the relation,
ν =c/ λ = (3 x 10⁸)/(9.7 x 10⁻⁸) ≈ 3.1 x 10¹⁵ Hz
Hence, the wavelength of the photon is 97 nm while the frequency is 3.1 x 10¹⁵ Hz.
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 xRead more
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.
Separation of two energy levels in an atom, E = 2.3 eV = 2.3 x 1.6 x 10⁻¹⁹ = 3.68 x 10⁻¹⁹J Let v 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 = Plank's constant = 6.62 x 10⁻34 Js Therefore , Read more
Separation of two energy levels in an atom,
E = 2.3 eV = 2.3 x 1.6 x 10⁻¹⁹ = 3.68 x 10⁻¹⁹J
Let v 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 = Plank’s constant = 6.62 x 10⁻34 Js
Therefore , ν= E/h = (3.68 x 10⁻¹⁹)/( 6.62 x 10⁻34) = 5.55 x 1014 Hz Hence, the frequency of the radiation is 5.6 x 1014 Hz.
Rydberg's formula is given as: hc /λ = 21.76 x 10⁻¹⁹ [1/n²₁ -1/n²₂] Where, h = Planck's constant = 6.6 x 10⁻34 Js and c = Speed of light = 3 x 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=∝ hc /λ = 21Read more
Rydberg’s formula is given as:
hc /λ = 21.76 x 10⁻¹⁹ [1/n²₁ -1/n²₂]
Where,
h = Planck’s constant = 6.6 x 10⁻34 Js and c = Speed of light = 3 x 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=∝
hc /λ = 21.76 x 10⁻¹⁹ [1/(3)² -1/(∝) ²]
λ = (6.6 x 10⁻34 ) x (3 x 10⁸) x 9 ]/( 21.76 x 10⁻¹⁹)
(a) Using the Bohr’s model calculate the speed of the electron in a hydrogen atom in the n = 1, 2, and 3 levels. (b) Calculate the orbital period in each of these levels.
Ans (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, ν1 = e²/n1 4π ε0 (h/2π) = e²/2 n1ε0h Where, e = 1.6 x 10⁻19 C ε0 = Permittivity of free space = 8.85 x 10⁻12 N-1 C2 m2 h = Planck’s cRead more
Ans (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,
ν1 = e²/n1 4π ε0 (h/2π) = e²/2 n1ε0h
Where, e = 1.6 x 10⁻19 C
ε0 = Permittivity of free space = 8.85 x 10⁻12 N-1 C2 m2
h = Planck’s constant = 6.62 x 10⁻³⁴ Js
Therefore, ν1 = (1.6 x 10⁻19 )2 /2 x (8.85 x 10⁻12) x (6.62 x 10⁻³⁴)
= 0.0218 x 10⁸ = 2.18 x 10⁶ m/s
For level n2= 2 ,we can write the relation for the corresponding orbital speed as:
ν2 = = e²/2 n2ε0h = (1.6 x 10⁻19 )2 /2 x 2 x(8.85 x 10⁻12) x (6.62 x 10⁻³⁴)
=1.09 x 10⁶ m/s
For level n3= 3 ,we can write the relation for the corresponding orbital speed as:
ν3 = = e²/2 n3ε0h = (1.6 x 10⁻19 )2 /2 x 3 x (8.85 x 10⁻12) x (6.62 x 10⁻³⁴)
=7.27 x 10⁵ m/s
Hence, the speed of the electron in a hydrogen atom in n = 1, n=2, and n=3 is 2.18 x 106 m/s, 1.09 x 106 m/s, 7.27 x 105 m/s respectively.
Ans (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:
T1 = 2π r/ν1
Where,r1 = Radius of the orbit =( n1)²h²ε0/πme²
h = Planck’s constant = 6.62 x 10⁻³⁴ Js and
e = Charge on an electron = 1.6 x 10⁻19 C.
ε0 = Permittivity of free space = 8.85 x 10⁻12 N⁻1C2 m⁻²
m = Mass of an electron = 9.1 x 10⁻31 kg
Therefore, T1 = 2π r1/ν1
= 2π x (1)² x (6.62 x 10⁻³⁴)² x (8.85 x 10⁻12)/[(2.18 x 10⁶) x π x ( 9.1 x 10⁻31) x (1.6 x 10⁻19)² ]
= 15.27 x 10⁻¹⁷ = 1.527 x 10⁻¹⁶ s
For level n2 = 2, we can write the period as:
T2 = 2π r2/ν2
Where, r2 = Radius of the electron in n2 = 2
= (n2)²h²ε0/πme²
Therefore,T2 =2π r2/ν2
=2π x (2)² x (6.62 x 10⁻³⁴)² x (8.85 x 10⁻12)/[(1.09 x 10⁶) x π x ( 9.1 x 10⁻31) x (1.6 x 10⁻19)² ]
=1.22 x 10⁻¹⁵ s
For level n3 = 3, we can write the period as:
T3 = 2π r3/ν3
Where, r3 = Radius of the electron in n3 = 3
= (n3)²h²ε0/πme²
Therefore,T3 =2π r3/ν3
=2π x (3)² x (6.62 x 10⁻³⁴)² x (8.85 x 10⁻12)/[(7.27 x 10⁵) x π x ( 9.1 x 10⁻31) x (1.6 x 10⁻19)² ]
=4.12 x 10⁻¹⁵ s
Hence, the orbital period in each of these levels is 1.527 x 10⁻¹⁶, 1.22 x 10⁻¹⁵ s, and 4.12 x 10⁻¹⁵ s respectively.
See lessA hydrogen atom initially in the ground level absorbs a photon, which excites it to the n = 4 level. Determine the wavelength and frequency of photon.
For ground level, n₁ = 1 Let E₁be the energy of this level. It is known that E₁ is related with m as: E₁ =-13.6/n²₁ eV = -13.6 /1² = -13.6 eV 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: Therefore E2 =-13.Read more
For ground level, n₁ = 1
Let E₁be the energy of this level. It is known that E₁ is related with m as:
E₁ =-13.6/n²₁ eV
= -13.6 /1² = -13.6 eV
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:
Therefore E2 =-13.6/n²2 eV
=-13.6/4² = -13.6 /16 eV
The amount of energy absorbed by the photon is given as :
E = E2-E1
= -13.6 /16 -(-13.6 )
= 13.6 x 15/16 eV
= 13.6 x 15/16 x 1.6 x 10⁻¹⁹ = 2.04 x 10⁻¹⁸ J
For a photon of wavelength λ, the expression of energy is written as:
E = hc/λ
Where,
h = Planck’s constant = 6.6 x 10⁻34 Js and c = Speed of light = 3 x 10⁸ m/s.
Therefore , λ = hc/E = (6.6 x 10⁻34) x (3 x 10⁸ ) /(2.04 x 10⁻¹⁸ )
= 9.7 x 10⁻⁸ m = 97nm
And, frequency of a photon is given by the relation,
ν =c/ λ = (3 x 10⁸)/(9.7 x 10⁻⁸) ≈ 3.1 x 10¹⁵ Hz
Hence, the wavelength of the photon is 97 nm while the frequency is 3.1 x 10¹⁵ Hz.
See lessThe ground state energy of hydrogen atom is –13.6 eV. What are the kinetic and potential energies of the electron in this state?
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 xRead more
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 x (13.6) = -27.2 eV
See lessA difference of 2.3 eV separates two energy levels in an atom. What is the frequency of radiation emitted when the atom make a transition from the upper level to the lower level?
Separation of two energy levels in an atom, E = 2.3 eV = 2.3 x 1.6 x 10⁻¹⁹ = 3.68 x 10⁻¹⁹J Let v 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 = Plank's constant = 6.62 x 10⁻34 Js Therefore , Read more
Separation of two energy levels in an atom,
E = 2.3 eV = 2.3 x 1.6 x 10⁻¹⁹ = 3.68 x 10⁻¹⁹J
Let v 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 = Plank’s constant = 6.62 x 10⁻34 Js
Therefore , ν= E/h = (3.68 x 10⁻¹⁹)/( 6.62 x 10⁻34) = 5.55 x 1014 Hz
See lessHence, the frequency of the radiation is 5.6 x 1014 Hz.
What is the shortest wavelength present in the Paschen series of spectral lines?
Rydberg's formula is given as: hc /λ = 21.76 x 10⁻¹⁹ [1/n²₁ -1/n²₂] Where, h = Planck's constant = 6.6 x 10⁻34 Js and c = Speed of light = 3 x 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=∝ hc /λ = 21Read more
Rydberg’s formula is given as:
hc /λ = 21.76 x 10⁻¹⁹ [1/n²₁ -1/n²₂]
Where,
h = Planck’s constant = 6.6 x 10⁻34 Js and c = Speed of light = 3 x 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=∝
hc /λ = 21.76 x 10⁻¹⁹ [1/(3)² -1/(∝) ²]
λ = (6.6 x 10⁻34 ) x (3 x 10⁸) x 9 ]/( 21.76 x 10⁻¹⁹)
= 8.189 x 10⁻⁷m = 818.9 nn
See less