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What is the de Broglie wavelength of the α-particle accelerated through a potential difference V
- a)0.287/√V
- b)12.27/√V
- c)0.101/√V
- d)0.202/√V
Correct answer is option 'C'. Can you explain this answer?
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De Broglie Wavelength Calculation:
De Broglie wavelength is given by the formula: λ = h/p, where h is the Planck's constant and p is the momentum of the particle.
Given:
Potential difference, V = V
Charge of α-particle, q = 2e
Kinetic energy, K.E. = qV
Calculating Momentum:
The kinetic energy of the α-particle is given by: K.E. = 1/2mv²
Using the relation between kinetic energy and potential difference: 1/2mv² = qV
Substitute q = 2e and solve for v: v = √(2qV/m)
Momentum, p = mv = m√(2qV/m) = √(2mqV)
Calculating De Broglie Wavelength:
De Broglie wavelength, λ = h/p = h/(√(2mqV))
Substitute the values of h, m, q, and V to get: λ = h/(√(2*mass of α-particle*charge*potential difference))
λ = h/(√(2*4*mass of electron*potential difference))
λ = h/(√(8*mass of electron*potential difference))
λ = h/(√(8*9.11x10^-31*potential difference))
λ = h/(√(72.88x10^-31*potential difference))
λ = h/(√(7.288x10^-30*potential difference))
λ = h/(0.101√potential difference)
Therefore, the de Broglie wavelength of the α-particle accelerated through a potential difference V is 0.101/√V. Hence, the correct answer is option 'C'.
De Broglie wavelength is given by the formula: λ = h/p, where h is the Planck's constant and p is the momentum of the particle.
Given:
Potential difference, V = V
Charge of α-particle, q = 2e
Kinetic energy, K.E. = qV
Calculating Momentum:
The kinetic energy of the α-particle is given by: K.E. = 1/2mv²
Using the relation between kinetic energy and potential difference: 1/2mv² = qV
Substitute q = 2e and solve for v: v = √(2qV/m)
Momentum, p = mv = m√(2qV/m) = √(2mqV)
Calculating De Broglie Wavelength:
De Broglie wavelength, λ = h/p = h/(√(2mqV))
Substitute the values of h, m, q, and V to get: λ = h/(√(2*mass of α-particle*charge*potential difference))
λ = h/(√(2*4*mass of electron*potential difference))
λ = h/(√(8*mass of electron*potential difference))
λ = h/(√(8*9.11x10^-31*potential difference))
λ = h/(√(72.88x10^-31*potential difference))
λ = h/(√(7.288x10^-30*potential difference))
λ = h/(0.101√potential difference)
Therefore, the de Broglie wavelength of the α-particle accelerated through a potential difference V is 0.101/√V. Hence, the correct answer is option 'C'.
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What is the de Broglie wavelength of the α-particle accelerated through a potential difference Va)0.287/√Vb)12.27/√Vc)0.101/√Vd)0.202/√VCorrect answer is option 'C'. Can you explain this answer?
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