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The de Broglie wavelength of a relativistic electron having 1 MeV of energy is ________× 10-12 m. (Take the rest mass energy of the electron to be 0.5 MeV. Planck constant = 6.63×10-34 Js, Speed of light= 3×108 m/s, Electronic charge = 1.6×10-19C)
Correct answer is between '1.3,1.6'. Can you explain this answer?
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The de Broglie wavelength of a relativistic electron having 1 MeV of e...
**Given information:**
- Energy of the relativistic electron: 1 MeV
- Rest mass energy of the electron: 0.5 MeV
- Planck constant: 6.63 × 10^-34 Js
- Speed of light: 3 × 10^8 m/s
- Electronic charge: 1.6 × 10^-19 C
**Calculating the momentum of the electron:**
The energy of a relativistic particle can be calculated using the equation:
E^2 = (pc)^2 + (mc^2)^2
where E is the total energy, p is the momentum, m is the rest mass, and c is the speed of light.
Rearranging the equation, we get:
p = sqrt(E^2 - (mc^2)^2) / c
Substituting the given values, we have:
p = sqrt((1 MeV)^2 - (0.5 MeV)^2) / (3 × 10^8 m/s)
p = sqrt((1 × 10^6 eV)^2 - (0.5 × 10^6 eV)^2) / (3 × 10^8 m/s)
p = sqrt((10^12 eV^2 - 0.25 × 10^12 eV^2) / (3 × 10^16 m^2/s^2)
p = sqrt(0.75 × 10^12 eV^2) / (3 × 10^8 m/s)
p = (0.866 × 10^6 eV) / (3 × 10^8 m/s)
p ≈ 2.887 × 10^-3 eV s/m
**Calculating the de Broglie wavelength:**
The de Broglie wavelength is given by the equation:
λ = h / p
where λ is the wavelength, h is the Planck constant, and p is the momentum.
Substituting the given values, we have:
λ = (6.63 × 10^-34 Js) / (2.887 × 10^-3 eV s/m)
λ = (6.63 × 10^-34 Js) / (2.887 × 10^-3 × 1.6 × 10^-19 C s/m)
λ = (6.63 × 10^-34 Js) / (4.6192 × 10^-22 C s/m)
λ ≈ 1.432 × 10^-12 m
Therefore, the de Broglie wavelength of the relativistic electron is approximately 1.432 × 10^-12 m.
Since the correct answer range is between 1.3 and 1.6, the calculated value falls within this range, confirming the accuracy of the solution.
- Energy of the relativistic electron: 1 MeV
- Rest mass energy of the electron: 0.5 MeV
- Planck constant: 6.63 × 10^-34 Js
- Speed of light: 3 × 10^8 m/s
- Electronic charge: 1.6 × 10^-19 C
**Calculating the momentum of the electron:**
The energy of a relativistic particle can be calculated using the equation:
E^2 = (pc)^2 + (mc^2)^2
where E is the total energy, p is the momentum, m is the rest mass, and c is the speed of light.
Rearranging the equation, we get:
p = sqrt(E^2 - (mc^2)^2) / c
Substituting the given values, we have:
p = sqrt((1 MeV)^2 - (0.5 MeV)^2) / (3 × 10^8 m/s)
p = sqrt((1 × 10^6 eV)^2 - (0.5 × 10^6 eV)^2) / (3 × 10^8 m/s)
p = sqrt((10^12 eV^2 - 0.25 × 10^12 eV^2) / (3 × 10^16 m^2/s^2)
p = sqrt(0.75 × 10^12 eV^2) / (3 × 10^8 m/s)
p = (0.866 × 10^6 eV) / (3 × 10^8 m/s)
p ≈ 2.887 × 10^-3 eV s/m
**Calculating the de Broglie wavelength:**
The de Broglie wavelength is given by the equation:
λ = h / p
where λ is the wavelength, h is the Planck constant, and p is the momentum.
Substituting the given values, we have:
λ = (6.63 × 10^-34 Js) / (2.887 × 10^-3 eV s/m)
λ = (6.63 × 10^-34 Js) / (2.887 × 10^-3 × 1.6 × 10^-19 C s/m)
λ = (6.63 × 10^-34 Js) / (4.6192 × 10^-22 C s/m)
λ ≈ 1.432 × 10^-12 m
Therefore, the de Broglie wavelength of the relativistic electron is approximately 1.432 × 10^-12 m.
Since the correct answer range is between 1.3 and 1.6, the calculated value falls within this range, confirming the accuracy of the solution.
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The de Broglie wavelength of a relativistic electron having 1 MeV of energy is________ 10-12 m. (Take the rest mass energy of the electron to be 0.5 MeV. Planck constant = 6.6310-34 Js, Speed of light= 3108 m/s, Electronic charge = 1.610-19C)Correct answer is between '1.3,1.6'. Can you explain this answer?
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The de Broglie wavelength of a relativistic electron having 1 MeV of energy is________ 10-12 m. (Take the rest mass energy of the electron to be 0.5 MeV. Planck constant = 6.6310-34 Js, Speed of light= 3108 m/s, Electronic charge = 1.610-19C)Correct answer is between '1.3,1.6'. Can you explain this answer? for IIT JAM 2024 is part of IIT JAM preparation. The Question and answers have been prepared according to the IIT JAM exam syllabus. Information about The de Broglie wavelength of a relativistic electron having 1 MeV of energy is________ 10-12 m. (Take the rest mass energy of the electron to be 0.5 MeV. Planck constant = 6.6310-34 Js, Speed of light= 3108 m/s, Electronic charge = 1.610-19C)Correct answer is between '1.3,1.6'. Can you explain this answer? covers all topics & solutions for IIT JAM 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The de Broglie wavelength of a relativistic electron having 1 MeV of energy is________ 10-12 m. (Take the rest mass energy of the electron to be 0.5 MeV. Planck constant = 6.6310-34 Js, Speed of light= 3108 m/s, Electronic charge = 1.610-19C)Correct answer is between '1.3,1.6'. Can you explain this answer?.
The de Broglie wavelength of a relativistic electron having 1 MeV of energy is________ 10-12 m. (Take the rest mass energy of the electron to be 0.5 MeV. Planck constant = 6.6310-34 Js, Speed of light= 3108 m/s, Electronic charge = 1.610-19C)Correct answer is between '1.3,1.6'. Can you explain this answer? for IIT JAM 2024 is part of IIT JAM preparation. The Question and answers have been prepared according to the IIT JAM exam syllabus. Information about The de Broglie wavelength of a relativistic electron having 1 MeV of energy is________ 10-12 m. (Take the rest mass energy of the electron to be 0.5 MeV. Planck constant = 6.6310-34 Js, Speed of light= 3108 m/s, Electronic charge = 1.610-19C)Correct answer is between '1.3,1.6'. Can you explain this answer? covers all topics & solutions for IIT JAM 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The de Broglie wavelength of a relativistic electron having 1 MeV of energy is________ 10-12 m. (Take the rest mass energy of the electron to be 0.5 MeV. Planck constant = 6.6310-34 Js, Speed of light= 3108 m/s, Electronic charge = 1.610-19C)Correct answer is between '1.3,1.6'. Can you explain this answer?.
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