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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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de-Brogile wavelength associated with a proton accelerated from rest b...
, where k is kinetic energy of particle. According to given data, kinetic energy of proton is 10keV which is same as that of neutron. Masses are also same. Most Upvoted Answer
de-Brogile wavelength associated with a proton accelerated from rest b...
To calculate the de Broglie wavelength associated with a proton accelerated from rest by 10 kV, we can use the equation:
λ = h / p
where λ is the de Broglie wavelength, h is the Planck's constant (6.626 x 10^-34 J·s), and p is the momentum of the proton.
The momentum of a particle can be calculated using the equation:
p = √(2mE)
where m is the mass of the proton (1.67 x 10^-27 kg) and E is the energy of the proton.
Given that the proton is accelerated from rest by 10 kV, we can convert this energy to joules:
E = 10,000 volts = 10,000 J/C * 1 electron / 1.6 x 10^-19 C = 6.25 x 10^-15 J
Now we can calculate the momentum of the proton:
p = √(2 * 1.67 x 10^-27 kg * 6.25 x 10^-15 J) = 5.83 x 10^-21 kg·m/s
Finally, we can calculate the de Broglie wavelength:
λ = 6.626 x 10^-34 J·s / 5.83 x 10^-21 kg·m/s ≈ 1.13 x 10^-13 meters
Therefore, the de Broglie wavelength associated with a proton accelerated from rest by 10 kV is approximately 1.13 x 10^-13 meters.
λ = h / p
where λ is the de Broglie wavelength, h is the Planck's constant (6.626 x 10^-34 J·s), and p is the momentum of the proton.
The momentum of a particle can be calculated using the equation:
p = √(2mE)
where m is the mass of the proton (1.67 x 10^-27 kg) and E is the energy of the proton.
Given that the proton is accelerated from rest by 10 kV, we can convert this energy to joules:
E = 10,000 volts = 10,000 J/C * 1 electron / 1.6 x 10^-19 C = 6.25 x 10^-15 J
Now we can calculate the momentum of the proton:
p = √(2 * 1.67 x 10^-27 kg * 6.25 x 10^-15 J) = 5.83 x 10^-21 kg·m/s
Finally, we can calculate the de Broglie wavelength:
λ = 6.626 x 10^-34 J·s / 5.83 x 10^-21 kg·m/s ≈ 1.13 x 10^-13 meters
Therefore, the de Broglie wavelength associated with a proton accelerated from rest by 10 kV is approximately 1.13 x 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 isa)λ / 2b)λc)2λd)2λ / 3Correct answer is option 'B'. Can you explain this answer?
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