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de-Brogile wavelength associated with a proton accelerated from rest by 10 kV is λ. The de-Brogile wavelength associated with a 10 keV neutron is
- a)λ/2
- b)λ
- c)2λ
- d)2λ/3
Correct answer is option 'B'. Can you explain this answer?
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To calculate the de Broglie wavelength of a proton accelerated from rest by 10 kV, we can use the following formula:
λ = h / p
Where:
- λ is the de Broglie wavelength
- h is the Planck's constant (6.62607015 × 10^-34 J⋅s)
- p is the momentum of the proton
The momentum of a proton can be calculated using the following equation:
p = √(2mE)
Where:
- p is the momentum of the proton
- m is the mass of the proton (1.67 × 10^-27 kg)
- E is the kinetic energy of the proton
Given that the proton is accelerated from rest by 10 kV, we can calculate the kinetic energy:
E = qV
Where:
- E is the kinetic energy
- q is the charge of the proton (1.6 × 10^-19 C)
- V is the voltage (10 kV = 10,000 V)
Now, let's calculate the kinetic energy:
E = (1.6 × 10^-19 C) * (10,000 V)
E = 1.6 × 10^-15 J
Next, we can calculate the momentum:
p = √(2 * (1.67 × 10^-27 kg) * (1.6 × 10^-15 J))
p ≈ 1.6 × 10^-21 kg⋅m/s
Finally, we can calculate the de Broglie wavelength:
λ = (6.62607015 × 10^-34 J⋅s) / (1.6 × 10^-21 kg⋅m/s)
λ ≈ 4.14 × 10^-13 m
Therefore, the de Broglie wavelength associated with a proton accelerated from rest by 10 kV is approximately 4.14 × 10^-13 meters.
λ = h / p
Where:
- λ is the de Broglie wavelength
- h is the Planck's constant (6.62607015 × 10^-34 J⋅s)
- p is the momentum of the proton
The momentum of a proton can be calculated using the following equation:
p = √(2mE)
Where:
- p is the momentum of the proton
- m is the mass of the proton (1.67 × 10^-27 kg)
- E is the kinetic energy of the proton
Given that the proton is accelerated from rest by 10 kV, we can calculate the kinetic energy:
E = qV
Where:
- E is the kinetic energy
- q is the charge of the proton (1.6 × 10^-19 C)
- V is the voltage (10 kV = 10,000 V)
Now, let's calculate the kinetic energy:
E = (1.6 × 10^-19 C) * (10,000 V)
E = 1.6 × 10^-15 J
Next, we can calculate the momentum:
p = √(2 * (1.67 × 10^-27 kg) * (1.6 × 10^-15 J))
p ≈ 1.6 × 10^-21 kg⋅m/s
Finally, we can calculate the de Broglie wavelength:
λ = (6.62607015 × 10^-34 J⋅s) / (1.6 × 10^-21 kg⋅m/s)
λ ≈ 4.14 × 10^-13 m
Therefore, the de Broglie wavelength associated with a proton accelerated from rest by 10 kV is approximately 4.14 × 10^-13 meters.
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de-Brogile wavelength associated with a proton accelerated from rest by 10 kV is λ. The de-Brogile wavelength associated with a 10 keV neutron is a)λ/2 b)λ c)2λd)2λ/3Correct answer is option 'B'. Can you explain this answer?
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