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The ratio of de-Broglie wavelength of a proton and an α-particle accelerated through the same potential difference is.
Correct answer is '2.828'. Can you explain this answer?
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The correct answer is: 2.828
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The correct answer is: 2.828
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The ratio of de-Broglie wavelength of a proton and anα-particle ...
Electron can be calculated using the formula:
λ = h / p
where λ is the de-Broglie wavelength, h is Planck's constant, and p is the momentum of the particle.
The momentum of a particle can be calculated using the formula:
p = mv
where m is the mass of the particle and v is its velocity.
The mass of a proton is approximately 1.6726219 x 10^-27 kg, and the mass of an electron is approximately 9.10938356 x 10^-31 kg.
The ratio of the de-Broglie wavelength of a proton to that of an electron can be calculated as follows:
λ_proton / λ_electron = (h / p_proton) / (h / p_electron)
Since h is common to both terms, it cancels out:
λ_proton / λ_electron = p_electron / p_proton
Using the formulas for momentum, we have:
λ_proton / λ_electron = (m_electron * v_electron) / (m_proton * v_proton)
Plugging in the values for the masses and velocities of the proton and electron, we get:
λ_proton / λ_electron = [(9.10938356 x 10^-31 kg) * v_electron] / [(1.6726219 x 10^-27 kg) * v_proton]
Simplifying further, we have:
λ_proton / λ_electron = 5.446 × 10^-4 * (v_electron / v_proton)
Therefore, the ratio of the de-Broglie wavelength of a proton to that of an electron is approximately 5.446 × 10^-4 times the ratio of their velocities.
λ = h / p
where λ is the de-Broglie wavelength, h is Planck's constant, and p is the momentum of the particle.
The momentum of a particle can be calculated using the formula:
p = mv
where m is the mass of the particle and v is its velocity.
The mass of a proton is approximately 1.6726219 x 10^-27 kg, and the mass of an electron is approximately 9.10938356 x 10^-31 kg.
The ratio of the de-Broglie wavelength of a proton to that of an electron can be calculated as follows:
λ_proton / λ_electron = (h / p_proton) / (h / p_electron)
Since h is common to both terms, it cancels out:
λ_proton / λ_electron = p_electron / p_proton
Using the formulas for momentum, we have:
λ_proton / λ_electron = (m_electron * v_electron) / (m_proton * v_proton)
Plugging in the values for the masses and velocities of the proton and electron, we get:
λ_proton / λ_electron = [(9.10938356 x 10^-31 kg) * v_electron] / [(1.6726219 x 10^-27 kg) * v_proton]
Simplifying further, we have:
λ_proton / λ_electron = 5.446 × 10^-4 * (v_electron / v_proton)
Therefore, the ratio of the de-Broglie wavelength of a proton to that of an electron is approximately 5.446 × 10^-4 times the ratio of their velocities.
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The ratio of de-Broglie wavelength of a proton and anα-particle accelerated through the same potential difference is.Correct answer is '2.828'. Can you explain this answer? for Physics 2025 is part of Physics preparation. The Question and answers have been prepared according to the Physics exam syllabus. Information about The ratio of de-Broglie wavelength of a proton and anα-particle accelerated through the same potential difference is.Correct answer is '2.828'. Can you explain this answer? covers all topics & solutions for Physics 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The ratio of de-Broglie wavelength of a proton and anα-particle accelerated through the same potential difference is.Correct answer is '2.828'. Can you explain this answer?.
The ratio of de-Broglie wavelength of a proton and anα-particle accelerated through the same potential difference is.Correct answer is '2.828'. Can you explain this answer? for Physics 2025 is part of Physics preparation. The Question and answers have been prepared according to the Physics exam syllabus. Information about The ratio of de-Broglie wavelength of a proton and anα-particle accelerated through the same potential difference is.Correct answer is '2.828'. Can you explain this answer? covers all topics & solutions for Physics 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The ratio of de-Broglie wavelength of a proton and anα-particle accelerated through the same potential difference is.Correct answer is '2.828'. Can you explain this answer?.
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