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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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What is the de Broglie wavelength of the α-particle accelerated throug...
De Broglie Wavelength of an α-particle accelerated through a potential difference
De Broglie Wavelength:
De Broglie wavelength is the wavelength associated with a moving particle, given by the equation λ = h/p, where λ is the wavelength, h is Planck's constant, and p is the momentum of the particle.
α-Particle Accelerated through a Potential Difference:
When an α-particle is accelerated through a potential difference, it gains kinetic energy, which can be calculated using the equation K = qV, where K is the kinetic energy, q is the charge of the α-particle, and V is the potential difference.
Calculating De Broglie Wavelength:
The momentum of the α-particle can be calculated using the equation p = √(2mK), where m is the mass of the α-particle. Therefore, the de Broglie wavelength can be calculated using the equation λ = h/√(2mK).
Given, V = 1.25 kV = 1250 V
Using the equation K = qV, we get K = (2)(1250) = 2500 eV
The mass of an α-particle is 6.64 x 10^-27 kg.
Substituting these values in the equation λ = h/√(2mK), we get λ = 6.63 x 10^-34 / √(2 x 6.64 x 10^-27 x 2500 x 1.6 x 10^-19) = 0.101 nm
Therefore, the correct answer is option C, which is 0.101/√V.
De Broglie Wavelength:
De Broglie wavelength is the wavelength associated with a moving particle, given by the equation λ = h/p, where λ is the wavelength, h is Planck's constant, and p is the momentum of the particle.
α-Particle Accelerated through a Potential Difference:
When an α-particle is accelerated through a potential difference, it gains kinetic energy, which can be calculated using the equation K = qV, where K is the kinetic energy, q is the charge of the α-particle, and V is the potential difference.
Calculating De Broglie Wavelength:
The momentum of the α-particle can be calculated using the equation p = √(2mK), where m is the mass of the α-particle. Therefore, the de Broglie wavelength can be calculated using the equation λ = h/√(2mK).
Given, V = 1.25 kV = 1250 V
Using the equation K = qV, we get K = (2)(1250) = 2500 eV
The mass of an α-particle is 6.64 x 10^-27 kg.
Substituting these values in the equation λ = h/√(2mK), we get λ = 6.63 x 10^-34 / √(2 x 6.64 x 10^-27 x 2500 x 1.6 x 10^-19) = 0.101 nm
Therefore, the correct answer is option C, which is 0.101/√V.
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