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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)
  • 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. 
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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.
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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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