The momentum of a photon having energy (hv) is given as: p = hv/c = h/λ => λ = h/p Where, λ = Wavelength of the electromagnetic radiation c = Speed of light h = Planck's constant De Broglie wavelength of the photon is given as:λ = h/mv But , p=mv, therefore λ = h/p Where, m = Mass of the photonRead more
The momentum of a photon having energy (hv) is given as:
p = hv/c = h/λ
=> λ = h/p
Where, λ = Wavelength of the electromagnetic radiation
c = Speed of light
h = Planck’s constant
De Broglie wavelength of the photon is given as:λ = h/mv
But , p=mv, therefore
λ = h/p
Where, m = Mass of the photon v = Velocity of the photon
Hence, it can be inferred from equations (i) and (ii) that the wavelength of the electromagnetic radiation is equal to the de Broglie wavelength of the photon.
Ans (a). De Broglie wavelength of the neutron, λ = 1.40 x 10⁻10 m Mass of a neutron, mn = 1.66 x 10-27 kg Planck's constant, h = 6.6 x 10⁻34 Js Kinetic energy (K) and velocity (v) are related as : K = 1/2 mnv2 -----------------Eq-1 De Broglie wavelength (λ) and velocity (v) are related as : λ = h/mnRead more
Ans (a).
De Broglie wavelength of the neutron, λ = 1.40 x 10⁻10 m
Mass of a neutron, mn = 1.66 x 10-27 kg
Planck’s constant, h = 6.6 x 10⁻34 Js
Kinetic energy (K) and velocity (v) are related as :
K = 1/2 mnv2 —————–Eq-1
De Broglie wavelength (λ) and velocity (v) are related as :
λ = h/mnv—————–Eq-2
Using Eq-2 in Eq-1 .we get :
K = 1/2 x (mnh²)/(λ²m²n) = h²/2 λ²mn
=(6.6 x 10⁻34)² /2 (1.40 x 10⁻10)² (1.66 x 10-27)
= 6.75 x 10⁻²¹ J
= (6.75 x 10⁻²¹) /(1.6 x 10⁻¹⁹ )
= 4.219 x 10⁻² eV
Hence ,the kinetic energy of the neutron is 6.75 x 10⁻²¹ J or 4.219 x 10⁻² eV
Ans (b).
Temperature of the neutron, T = 300 K
Boltzmann constant, k = 1.38 x 10-23 kg m2 s⁻2 K⁻1
Average kinetic energy of the neutron:
K’ = 3/2 k T = 3/2 x 1.38 x 10-23 x 300 = 6.21 x 10-21 J
The relation for the de Broglie wavelength is given as:
λ’ =h/ √ (2K’mn)
Where, mn = 1.66 x 10-27 kg, h = 6.6 x 10-34 Js and K‘ = 6.75 x 10-21 J.
Therefore,
λ’ = (6.6 x 10-34)/√[2 x (6.21 x 10-21) x (1.66 x 10-27) ]
= 1.46 x 10–10 m = 0.146 nm
Therefore, the de Broglie wavelength of the neutron is 0.146 nm.
Wavelength of an electron (λe) and a photon (λp), λe = λp = 1 nm = 1 x 10⁻9 m Planck's constant, h = 6.63 x 10-34 Js Ans (a). The momentum of an elementary particle is given by de Broglie relation: λ = h/p => p = h/λ It is clear that momentum depends only on the wavelength of the particle. SincRead more
Wavelength of an electron (λe) and a photon (λp), λe = λp = 1 nm = 1 x 10⁻9 m
Planck’s constant, h = 6.63 x 10-34 Js
Ans (a).
The momentum of an elementary particle is given by de Broglie relation:
λ = h/p => p = h/λ
It is clear that momentum depends only on the wavelength of the particle. Since the wavelength of an electron and photon are equal, both have an equal momentum.
Therefore , p = 6.63 x 10⁻34 / ( 1 x 10⁻9)
= 6.63 x 10⁻²⁵ kg ms⁻¹
Ans (b).
The energy of a photon is given by the relation:
E = hc/λ
Where, Speed of light, c = 3 x 108 m/s ,
Therefore, E = (6.63 x 10⁻34 ) x (3 x 108 /(1 x 10⁻9) (1.6 x 10⁻¹⁹ )
=1243.1 eV = 1.243 kev
Therefore, the energy of the photon is 1.243 keV.
Ans (c).
The kinetic energy (K) of an electron having momentum p, is given by the relation:
K = 1/2 p²/m
Where,
m = Mass of the electron = 9.1 x 10-31 kg
p = 6.63 x 10-25 kg m s-1
Therefore , K = 1/2 ( 6.63 x 10-25 )² /( 9.1 x 10-31) = 2.415 x 10⁻¹⁹ J
= (2.415 x 10⁻¹⁹) /(1.6 x 10⁻¹⁹) = 1.51 eV
Hence, the kinetic energy of the electron is 1.51 eV.
Ans (a). Mass of the bullet, m = 0.040 kg Speed of the bullet, v = 1.0 km/s = 1000 m/s Planck’s constant, h = 6.6 x 10⁻34 Js De Broglie wavelength of the bullet is given by the relation: λ=h/mv =( 6.6 x 10⁻34)/(0.040) (1000) = 1.65 x 10 ⁻³⁵ m Ans (b). Mass of the ball, m = 0.060 kg Speed of the ballRead more
Ans (a).
Mass of the bullet, m = 0.040 kg
Speed of the bullet, v = 1.0 km/s = 1000 m/s
Planck’s constant, h = 6.6 x 10⁻34 Js
De Broglie wavelength of the bullet is given by the relation:
λ=h/mv
=( 6.6 x 10⁻34)/(0.040) (1000) = 1.65 x 10 ⁻³⁵ m
Ans (b).
Mass of the ball, m = 0.060 kg
Speed of the ball, v = 1.0 m/s
De Broglie wavelength of the ball is given by the relation:
λ=h/mv
=( 6.6 x 10⁻34)/(0.060) (1) = 1.1 x 10 ⁻³² m
Ans (c).
Mass of the dust particle, m = 1 x 10-9 kg
Speed of the dust particle, v = 2.2 m/s
De Broglie wavelength of the dust particle is given by the relation:
Wavelength of light of a sodium line, λ = 589 nm = 589 x 10-9 m Mass of an electron, me= 9.1 x 10-31 kg Mass of a neutron, mn= 1.66 x 10-27 kg Planck's constant, h = 6.6 x 10-34 Js Ans (a). For the kinetic energy K, of an electron accelerating with a velocity v, we have the relation: We have the relRead more
Wavelength of light of a sodium line, λ = 589 nm = 589 x 10–9 m
Mass of an electron, me= 9.1 x 10–31 kg
Mass of a neutron, mn= 1.66 x 10–27 kg
Planck’s constant, h = 6.6 x 10-34 Js
Ans (a).
For the kinetic energy K, of an electron accelerating with a velocity v, we have the relation:
We have the relation for de Broglie wavelength as:
K = 1/2 mev²—————-Eq-1
We have the relation for de Broglie wavelength as :
λ = h /mev
Therefore, v² =h²/λ²m²e——————— Eq-2 Substituting equation (2) in equation (1), we get the relation:
K = 1/2 meh²/2λ²m²e =h²/2λ²me—————-Eq-3
= (6.6 x 10-34)²/2 (589 x 10–9)² (9.1 x 10–31 )
≈ 6.9 x 10-25 J
=(6.9 x 10-25)/(1.6 x 10⁻¹⁹ ) =4.31 x 10 ⁻⁶ eV = 4.31 μ eV
Hence, the kinetic energy of the electron is 6.9 x 10-25 J or 4.31 μeV.
Ans (b).
Using equation (3), we can write the relation for the kinetic energy of the neutron as:
K= h²/2 λ²mn
= (6.6 x 10-34)² /2 (589 x 10–9)² (1.66 x 10–27 )
=3.78 x 10⁻²⁸ J
=( 3.78 x 10⁻²⁸ )/(1.6 x 10⁻¹⁹ )
= 2.36 x 10⁻⁹ eV = 2.36 neV
Hence, the kinetic energy of the neutron is 3.78 x 10⁻28 J or 2.36 neV.
Show that the wavelength of electromagnetic radiation is equal to the de Broglie wavelength of its quantum (photon).
The momentum of a photon having energy (hv) is given as: p = hv/c = h/λ => λ = h/p Where, λ = Wavelength of the electromagnetic radiation c = Speed of light h = Planck's constant De Broglie wavelength of the photon is given as:λ = h/mv But , p=mv, therefore λ = h/p Where, m = Mass of the photonRead more
The momentum of a photon having energy (hv) is given as:
p = hv/c = h/λ
=> λ = h/p
Where, λ = Wavelength of the electromagnetic radiation
c = Speed of light
h = Planck’s constant
De Broglie wavelength of the photon is given as:λ = h/mv
But , p=mv, therefore
λ = h/p
Where, m = Mass of the photon v = Velocity of the photon
Hence, it can be inferred from equations (i) and (ii) that the wavelength of the electromagnetic radiation is equal to the de Broglie wavelength of the photon.
See less(a) For what kinetic energy of a neutron will the associated de Broglie wavelength be 1.40 × 10⁻¹⁰ m? (b) Also find the de Broglie wavelength of a neutron, in thermal equilibrium with matter, having an average kinetic energy of (3/2) k T at 300 K
Ans (a). De Broglie wavelength of the neutron, λ = 1.40 x 10⁻10 m Mass of a neutron, mn = 1.66 x 10-27 kg Planck's constant, h = 6.6 x 10⁻34 Js Kinetic energy (K) and velocity (v) are related as : K = 1/2 mnv2 -----------------Eq-1 De Broglie wavelength (λ) and velocity (v) are related as : λ = h/mnRead more
Ans (a).
De Broglie wavelength of the neutron, λ = 1.40 x 10⁻10 m
Mass of a neutron, mn = 1.66 x 10-27 kg
Planck’s constant, h = 6.6 x 10⁻34 Js
Kinetic energy (K) and velocity (v) are related as :
K = 1/2 mnv2 —————–Eq-1
De Broglie wavelength (λ) and velocity (v) are related as :
λ = h/mnv—————–Eq-2
Using Eq-2 in Eq-1 .we get :
K = 1/2 x (mnh²)/(λ²m²n) = h²/2 λ²mn
=(6.6 x 10⁻34)² /2 (1.40 x 10⁻10)² (1.66 x 10-27)
= 6.75 x 10⁻²¹ J
= (6.75 x 10⁻²¹) /(1.6 x 10⁻¹⁹ )
= 4.219 x 10⁻² eV
Hence ,the kinetic energy of the neutron is 6.75 x 10⁻²¹ J or 4.219 x 10⁻² eV
Ans (b).
Temperature of the neutron, T = 300 K
Boltzmann constant, k = 1.38 x 10-23 kg m2 s⁻2 K⁻1
Average kinetic energy of the neutron:
K’ = 3/2 k T = 3/2 x 1.38 x 10-23 x 300 = 6.21 x 10-21 J
The relation for the de Broglie wavelength is given as:
λ’ =h/ √ (2K’mn)
Where, mn = 1.66 x 10-27 kg, h = 6.6 x 10-34 Js and K‘ = 6.75 x 10-21 J.
Therefore,
λ’ = (6.6 x 10-34)/√[2 x (6.21 x 10-21) x (1.66 x 10-27) ]
= 1.46 x 10–10 m = 0.146 nm
Therefore, the de Broglie wavelength of the neutron is 0.146 nm.
See lessAn electron and a photon each have a wavelength of 1.00 nm. Find (a) their momenta, (b) the energy of the photon, and (c) the kinetic energy of electron
Wavelength of an electron (λe) and a photon (λp), λe = λp = 1 nm = 1 x 10⁻9 m Planck's constant, h = 6.63 x 10-34 Js Ans (a). The momentum of an elementary particle is given by de Broglie relation: λ = h/p => p = h/λ It is clear that momentum depends only on the wavelength of the particle. SincRead more
Wavelength of an electron (λe) and a photon (λp), λe = λp = 1 nm = 1 x 10⁻9 m
Planck’s constant, h = 6.63 x 10-34 Js
Ans (a).
The momentum of an elementary particle is given by de Broglie relation:
λ = h/p => p = h/λ
It is clear that momentum depends only on the wavelength of the particle. Since the wavelength of an electron and photon are equal, both have an equal momentum.
Therefore , p = 6.63 x 10⁻34 / ( 1 x 10⁻9)
= 6.63 x 10⁻²⁵ kg ms⁻¹
Ans (b).
The energy of a photon is given by the relation:
E = hc/λ
Where, Speed of light, c = 3 x 108 m/s ,
Therefore, E = (6.63 x 10⁻34 ) x (3 x 108 /(1 x 10⁻9) (1.6 x 10⁻¹⁹ )
=1243.1 eV = 1.243 kev
Therefore, the energy of the photon is 1.243 keV.
Ans (c).
The kinetic energy (K) of an electron having momentum p, is given by the relation:
K = 1/2 p²/m
Where,
m = Mass of the electron = 9.1 x 10-31 kg
p = 6.63 x 10-25 kg m s-1
Therefore , K = 1/2 ( 6.63 x 10-25 )² /( 9.1 x 10-31) = 2.415 x 10⁻¹⁹ J
= (2.415 x 10⁻¹⁹) /(1.6 x 10⁻¹⁹) = 1.51 eV
Hence, the kinetic energy of the electron is 1.51 eV.
See lessWhat is the de Broglie wavelength of (a) a bullet of mass 0.040 kg travelling at the speed of 1.0 km/s, (b) a ball of mass 0.060 kg moving at a speed of 1.0 m/s, and (c) a dust particle of mass 1.0 × 10⁻⁹ kg drifting with a speed of 2.2 m/s?
Ans (a). Mass of the bullet, m = 0.040 kg Speed of the bullet, v = 1.0 km/s = 1000 m/s Planck’s constant, h = 6.6 x 10⁻34 Js De Broglie wavelength of the bullet is given by the relation: λ=h/mv =( 6.6 x 10⁻34)/(0.040) (1000) = 1.65 x 10 ⁻³⁵ m Ans (b). Mass of the ball, m = 0.060 kg Speed of the ballRead more
Ans (a).
Mass of the bullet, m = 0.040 kg
Speed of the bullet, v = 1.0 km/s = 1000 m/s
Planck’s constant, h = 6.6 x 10⁻34 Js
De Broglie wavelength of the bullet is given by the relation:
λ=h/mv
=( 6.6 x 10⁻34)/(0.040) (1000) = 1.65 x 10 ⁻³⁵ m
Ans (b).
Mass of the ball, m = 0.060 kg
Speed of the ball, v = 1.0 m/s
De Broglie wavelength of the ball is given by the relation:
λ=h/mv
=( 6.6 x 10⁻34)/(0.060) (1) = 1.1 x 10 ⁻³² m
Ans (c).
Mass of the dust particle, m = 1 x 10-9 kg
Speed of the dust particle, v = 2.2 m/s
De Broglie wavelength of the dust particle is given by the relation:
λ=h/mv
=( 6.6 x 10⁻34)/(2.2) (1 x 10-9) = 3.0 x 10 ⁻²⁵ m
See lessThe wavelength of light from the spectral emission line of sodium is 589 nm. Find the kinetic energy at which (a) an electron, and (b) a neutron, would have the same de Broglie wavelength.
Wavelength of light of a sodium line, λ = 589 nm = 589 x 10-9 m Mass of an electron, me= 9.1 x 10-31 kg Mass of a neutron, mn= 1.66 x 10-27 kg Planck's constant, h = 6.6 x 10-34 Js Ans (a). For the kinetic energy K, of an electron accelerating with a velocity v, we have the relation: We have the relRead more
Wavelength of light of a sodium line, λ = 589 nm = 589 x 10–9 m
Mass of an electron, me= 9.1 x 10–31 kg
Mass of a neutron, mn= 1.66 x 10–27 kg
Planck’s constant, h = 6.6 x 10-34 Js
Ans (a).
For the kinetic energy K, of an electron accelerating with a velocity v, we have the relation:
We have the relation for de Broglie wavelength as:
K = 1/2 mev²—————-Eq-1
We have the relation for de Broglie wavelength as :
λ = h /mev
Therefore, v² =h²/λ²m²e——————— Eq-2
Substituting equation (2) in equation (1), we get the relation:
K = 1/2 meh²/2λ²m²e =h²/2λ²me—————-Eq-3
= (6.6 x 10-34)²/2 (589 x 10–9)² (9.1 x 10–31 )
≈ 6.9 x 10-25 J
=(6.9 x 10-25)/(1.6 x 10⁻¹⁹ ) =4.31 x 10 ⁻⁶ eV = 4.31 μ eV
Hence, the kinetic energy of the electron is 6.9 x 10-25 J or 4.31 μeV.
Ans (b).
Using equation (3), we can write the relation for the kinetic energy of the neutron as:
K= h²/2 λ²mn
= (6.6 x 10-34)² /2 (589 x 10–9)² (1.66 x 10–27 )
=3.78 x 10⁻²⁸ J
=( 3.78 x 10⁻²⁸ )/(1.6 x 10⁻¹⁹ )
= 2.36 x 10⁻⁹ eV = 2.36 neV
Hence, the kinetic energy of the neutron is 3.78 x 10⁻28 J or 2.36 neV.
See less