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Show that the de Broglie wavelength for a material particle of rest mass mo and charge q, accelerated from rest through a potential difference of V volts relativistically is given by, (Symbols has their usual meanings. Use relativistic energy equation) λ=h/√[2mqV(1 (qV/2mc²))]?
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Show that the de Broglie wavelength for a material particle of rest ma...
De Broglie Wavelength for a Material Particle
The de Broglie wavelength for a material particle can be derived by considering the relativistic energy equation and the wave-particle duality of matter.
Relativistic Energy Equation
The relativistic energy equation relates the total energy (E) of a particle to its rest mass (mo) and its velocity (v). It is given by:
E = √[(mo*c²) + (pc)²]
Where c is the speed of light and p is the momentum of the particle.
Wave-Particle Duality
According to wave-particle duality, particles also exhibit wave-like properties. The wavelength (λ) associated with a particle is related to its momentum (p) by the de Broglie relation:
λ = h/p
Where h is the Planck's constant.
Derivation
1. Consider a material particle of rest mass mo and charge q, accelerated from rest through a potential difference of V volts.
2. The initial kinetic energy (K) of the particle can be calculated using the potential difference V as:
K = qV
3. According to the relativistic energy equation, the total energy (E) of the particle is given by:
E = √[(mo*c²) + (pc)²]
4. The momentum (p) of the particle can be calculated using the total energy (E) and the rest mass (mo) as:
p = √[(E² - (mo*c²)²)/c²]
5. Substituting the value of total energy (E) from step 3, we get:
p = √[(√[(mo*c²) + (pc)²]² - (mo*c²)²)/c²]
6. Simplifying the above equation, we get:
p = √[2moE - (mo²*c²)]
7. Substituting the value of kinetic energy (K) from step 2, we get:
p = √[2mo(qV) - (mo²*c²)]
8. Rearranging the equation, we get:
p = √[2moqV(1 - (qV/2mo*c²))]
9. Substituting the value of momentum (p) in the de Broglie relation, we get:
λ = h/p = h/√[2moqV(1 - (qV/2mo*c²))]
Conclusion
The de Broglie wavelength for a material particle of rest mass mo and charge q, accelerated from rest through a potential difference of V volts relativistically is given by the equation:
λ = h/√[2moqV(1 - (qV/2mo*c²))]
The de Broglie wavelength for a material particle can be derived by considering the relativistic energy equation and the wave-particle duality of matter.
Relativistic Energy Equation
The relativistic energy equation relates the total energy (E) of a particle to its rest mass (mo) and its velocity (v). It is given by:
E = √[(mo*c²) + (pc)²]
Where c is the speed of light and p is the momentum of the particle.
Wave-Particle Duality
According to wave-particle duality, particles also exhibit wave-like properties. The wavelength (λ) associated with a particle is related to its momentum (p) by the de Broglie relation:
λ = h/p
Where h is the Planck's constant.
Derivation
1. Consider a material particle of rest mass mo and charge q, accelerated from rest through a potential difference of V volts.
2. The initial kinetic energy (K) of the particle can be calculated using the potential difference V as:
K = qV
3. According to the relativistic energy equation, the total energy (E) of the particle is given by:
E = √[(mo*c²) + (pc)²]
4. The momentum (p) of the particle can be calculated using the total energy (E) and the rest mass (mo) as:
p = √[(E² - (mo*c²)²)/c²]
5. Substituting the value of total energy (E) from step 3, we get:
p = √[(√[(mo*c²) + (pc)²]² - (mo*c²)²)/c²]
6. Simplifying the above equation, we get:
p = √[2moE - (mo²*c²)]
7. Substituting the value of kinetic energy (K) from step 2, we get:
p = √[2mo(qV) - (mo²*c²)]
8. Rearranging the equation, we get:
p = √[2moqV(1 - (qV/2mo*c²))]
9. Substituting the value of momentum (p) in the de Broglie relation, we get:
λ = h/p = h/√[2moqV(1 - (qV/2mo*c²))]
Conclusion
The de Broglie wavelength for a material particle of rest mass mo and charge q, accelerated from rest through a potential difference of V volts relativistically is given by the equation:
λ = h/√[2moqV(1 - (qV/2mo*c²))]
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Show that the de Broglie wavelength for a material particle of rest mass mo and charge q, accelerated from rest through a potential difference of V volts relativistically is given by, (Symbols has their usual meanings. Use relativistic energy equation) λ=h/√[2mqV(1 (qV/2mc²))]?
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Show that the de Broglie wavelength for a material particle of rest mass mo and charge q, accelerated from rest through a potential difference of V volts relativistically is given by, (Symbols has their usual meanings. Use relativistic energy equation) λ=h/√[2mqV(1 (qV/2mc²))]? for Computer Science Engineering (CSE) 2024 is part of Computer Science Engineering (CSE) preparation. The Question and answers have been prepared according to the Computer Science Engineering (CSE) exam syllabus. Information about Show that the de Broglie wavelength for a material particle of rest mass mo and charge q, accelerated from rest through a potential difference of V volts relativistically is given by, (Symbols has their usual meanings. Use relativistic energy equation) λ=h/√[2mqV(1 (qV/2mc²))]? covers all topics & solutions for Computer Science Engineering (CSE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Show that the de Broglie wavelength for a material particle of rest mass mo and charge q, accelerated from rest through a potential difference of V volts relativistically is given by, (Symbols has their usual meanings. Use relativistic energy equation) λ=h/√[2mqV(1 (qV/2mc²))]?.
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