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A proton and an alpha particle are accelerated through a potential difference of 100 V. The ratio of the wavelength associated with the proton to that associated with an alpha particle is
- a)√2
- b)2
- c)2√2
- d)1/2√2
Correct answer is option 'C'. Can you explain this answer?
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A proton and an alpha particle are accelerated through a potential dif...
We can use the de Broglie wavelength formula to find the ratio of the wavelengths associated with the proton and the alpha particle:
λ = h/p
where λ is the wavelength, h is Planck's constant, and p is the momentum of the particle. The momentum of a particle can be expressed as:
p = sqrt(2mE)
where m is the mass of the particle, E is the kinetic energy gained by the particle due to the potential difference.
For the proton, m = 1.67 x 10^-27 kg and E = qV = 1.6 x 10^-19 C x 100 V = 1.6 x 10^-17 J.
So, the momentum of the proton is:
p = sqrt(2 x 1.67 x 10^-27 kg x 1.6 x 10^-17 J) = 2.20 x 10^-22 kg m/s
Therefore, the de Broglie wavelength of the proton is:
λ = h/p = 6.63 x 10^-34 J s / 2.20 x 10^-22 kg m/s = 3.01 x 10^-12 m
For the alpha particle, m = 6.64 x 10^-27 kg and E = qV = 2 x 1.6 x 10^-19 C x 100 V = 3.2 x 10^-17 J.
So, the momentum of the alpha particle is:
p = sqrt(2 x 6.64 x 10^-27 kg x 3.2 x 10^-17 J) = 1.57 x 10^-21 kg m/s
Therefore, the de Broglie wavelength of the alpha particle is:
λ = h/p = 6.63 x 10^-34 J s / 1.57 x 10^-21 kg m/s = 4.22 x 10^-14 m
The ratio of the wavelengths associated with the proton to that associated with the alpha particle is:
λ_proton / λ_alpha = (3.01 x 10^-12 m) / (4.22 x 10^-14 m) = 71.3
Therefore, the ratio of the wavelengths associated with the proton to that associated with the alpha particle is 71.3.
λ = h/p
where λ is the wavelength, h is Planck's constant, and p is the momentum of the particle. The momentum of a particle can be expressed as:
p = sqrt(2mE)
where m is the mass of the particle, E is the kinetic energy gained by the particle due to the potential difference.
For the proton, m = 1.67 x 10^-27 kg and E = qV = 1.6 x 10^-19 C x 100 V = 1.6 x 10^-17 J.
So, the momentum of the proton is:
p = sqrt(2 x 1.67 x 10^-27 kg x 1.6 x 10^-17 J) = 2.20 x 10^-22 kg m/s
Therefore, the de Broglie wavelength of the proton is:
λ = h/p = 6.63 x 10^-34 J s / 2.20 x 10^-22 kg m/s = 3.01 x 10^-12 m
For the alpha particle, m = 6.64 x 10^-27 kg and E = qV = 2 x 1.6 x 10^-19 C x 100 V = 3.2 x 10^-17 J.
So, the momentum of the alpha particle is:
p = sqrt(2 x 6.64 x 10^-27 kg x 3.2 x 10^-17 J) = 1.57 x 10^-21 kg m/s
Therefore, the de Broglie wavelength of the alpha particle is:
λ = h/p = 6.63 x 10^-34 J s / 1.57 x 10^-21 kg m/s = 4.22 x 10^-14 m
The ratio of the wavelengths associated with the proton to that associated with the alpha particle is:
λ_proton / λ_alpha = (3.01 x 10^-12 m) / (4.22 x 10^-14 m) = 71.3
Therefore, the ratio of the wavelengths associated with the proton to that associated with the alpha particle is 71.3.
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A proton and an alpha particle are accelerated through a potential difference of 100 V. The ratio of the wavelength associated with the proton to that associated with an alpha particle isa)√2b)2c)2√2d)1/2√2Correct answer is option 'C'. Can you explain this answer?
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