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Assuming Heisenberg Uncertainity Principle to be true what could be the miniumum uncertainity in de–Broglie wavelength of a moving electron accelerated by Potential Difference of 6V whose uncertainity in position is 7/22 n.m.
- a)6.25 Å
- b)6 Å
- c)0.625 Å
- d)0.3125 Å
Correct answer is option 'C'. Can you explain this answer?
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Solution:
Given: Potential difference, V = 6V
Uncertainty in position, Δx = 7/22 n.m.
The minimum uncertainty in de Broglie wavelength is given by the relation:
Δλ ≥ h/Δp
where, h is Planck's constant = 6.626 x 10^-34 J s
Δp is the uncertainty in momentum
The momentum of an electron accelerated through a potential difference, V is given by the relation:
p = √(2mV)
where, m is the mass of the electron = 9.11 x 10^-31 kg
Therefore, the uncertainty in momentum is given by:
Δp = √(2mΔV)
where, ΔV is the uncertainty in potential difference
The uncertainty in potential difference can be calculated using the formula:
ΔV = Δx/e
where, e is the charge of an electron = 1.6 x 10^-19 C
Substituting the given values, we get:
ΔV = (7/22 x 10^-9)/(1.6 x 10^-19) = 0.27375 V
Substituting these values in the expression for Δp, we get:
Δp = √(2 x 9.11 x 10^-31 x 0.27375) = 1.24 x 10^-24 kg m/s
Substituting the values of h and Δp in the expression for minimum uncertainty in de Broglie wavelength, we get:
Δλ ≥ h/Δp = 6.626 x 10^-34/1.24 x 10^-24 = 0.53467 x 10^-10 m
Converting to nanometers, we get:
Δλ ≥ 0.53467 x 10^-1 nm
Rounding off to one decimal place, we get:
Δλ ≥ 0.5 nm
Therefore, the minimum uncertainty in de Broglie wavelength is 0.5 nm, which is closest to option C (0.625 nm).
Given: Potential difference, V = 6V
Uncertainty in position, Δx = 7/22 n.m.
The minimum uncertainty in de Broglie wavelength is given by the relation:
Δλ ≥ h/Δp
where, h is Planck's constant = 6.626 x 10^-34 J s
Δp is the uncertainty in momentum
The momentum of an electron accelerated through a potential difference, V is given by the relation:
p = √(2mV)
where, m is the mass of the electron = 9.11 x 10^-31 kg
Therefore, the uncertainty in momentum is given by:
Δp = √(2mΔV)
where, ΔV is the uncertainty in potential difference
The uncertainty in potential difference can be calculated using the formula:
ΔV = Δx/e
where, e is the charge of an electron = 1.6 x 10^-19 C
Substituting the given values, we get:
ΔV = (7/22 x 10^-9)/(1.6 x 10^-19) = 0.27375 V
Substituting these values in the expression for Δp, we get:
Δp = √(2 x 9.11 x 10^-31 x 0.27375) = 1.24 x 10^-24 kg m/s
Substituting the values of h and Δp in the expression for minimum uncertainty in de Broglie wavelength, we get:
Δλ ≥ h/Δp = 6.626 x 10^-34/1.24 x 10^-24 = 0.53467 x 10^-10 m
Converting to nanometers, we get:
Δλ ≥ 0.53467 x 10^-1 nm
Rounding off to one decimal place, we get:
Δλ ≥ 0.5 nm
Therefore, the minimum uncertainty in de Broglie wavelength is 0.5 nm, which is closest to option C (0.625 nm).
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Assuming Heisenberg Uncertainity Principle to be true what could be the miniumum uncertainity in de–Broglie wavelength of a moving electron accelerated by Potential Difference of 6V whose uncertainity in position is 7/22 n.m.a)6.25 Åb)6 Åc)0.625 Åd)0.3125 ÅCorrect answer is option 'C'. Can you explain this answer?
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Assuming Heisenberg Uncertainity Principle to be true what could be the miniumum uncertainity in de–Broglie wavelength of a moving electron accelerated by Potential Difference of 6V whose uncertainity in position is 7/22 n.m.a)6.25 Åb)6 Åc)0.625 Åd)0.3125 ÅCorrect answer is option 'C'. Can you explain this answer? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about Assuming Heisenberg Uncertainity Principle to be true what could be the miniumum uncertainity in de–Broglie wavelength of a moving electron accelerated by Potential Difference of 6V whose uncertainity in position is 7/22 n.m.a)6.25 Åb)6 Åc)0.625 Åd)0.3125 ÅCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Assuming Heisenberg Uncertainity Principle to be true what could be the miniumum uncertainity in de–Broglie wavelength of a moving electron accelerated by Potential Difference of 6V whose uncertainity in position is 7/22 n.m.a)6.25 Åb)6 Åc)0.625 Åd)0.3125 ÅCorrect answer is option 'C'. Can you explain this answer?.
Assuming Heisenberg Uncertainity Principle to be true what could be the miniumum uncertainity in de–Broglie wavelength of a moving electron accelerated by Potential Difference of 6V whose uncertainity in position is 7/22 n.m.a)6.25 Åb)6 Åc)0.625 Åd)0.3125 ÅCorrect answer is option 'C'. Can you explain this answer? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about Assuming Heisenberg Uncertainity Principle to be true what could be the miniumum uncertainity in de–Broglie wavelength of a moving electron accelerated by Potential Difference of 6V whose uncertainity in position is 7/22 n.m.a)6.25 Åb)6 Åc)0.625 Åd)0.3125 ÅCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Assuming Heisenberg Uncertainity Principle to be true what could be the miniumum uncertainity in de–Broglie wavelength of a moving electron accelerated by Potential Difference of 6V whose uncertainity in position is 7/22 n.m.a)6.25 Åb)6 Åc)0.625 Åd)0.3125 ÅCorrect answer is option 'C'. Can you explain this answer?.
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Assuming Heisenberg Uncertainity Principle to be true what could be the miniumum uncertainity in de–Broglie wavelength of a moving electron accelerated by Potential Difference of 6V whose uncertainity in position is 7/22 n.m.a)6.25 Åb)6 Åc)0.625 Åd)0.3125 ÅCorrect answer is option 'C'. Can you explain this answer?, a detailed solution for Assuming Heisenberg Uncertainity Principle to be true what could be the miniumum uncertainity in de–Broglie wavelength of a moving electron accelerated by Potential Difference of 6V whose uncertainity in position is 7/22 n.m.a)6.25 Åb)6 Åc)0.625 Åd)0.3125 ÅCorrect answer is option 'C'. Can you explain this answer? has been provided alongside types of Assuming Heisenberg Uncertainity Principle to be true what could be the miniumum uncertainity in de–Broglie wavelength of a moving electron accelerated by Potential Difference of 6V whose uncertainity in position is 7/22 n.m.a)6.25 Åb)6 Åc)0.625 Åd)0.3125 ÅCorrect answer is option 'C'. Can you explain this answer? theory, EduRev gives you an
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