Class 11 Exam > Class 11 Questions > Assuming Heisenberg Uncertainity Principle to...
Start Learning for Free
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?
| FREE This question is part of | Download PDF Attempt this Test |
Verified Answer
Assuming Heisenberg Uncertainity Principle to be true what could be th...
Most Upvoted Answer
Assuming Heisenberg Uncertainity Principle to be true what could be th...
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).
Attention Class 11 Students!
To make sure you are not studying endlessly, EduRev has designed Class 11 study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in Class 11.
|
Explore Courses for Class 11 exam
|
|
Similar Class 11 Doubts
Top Courses for Class 11View all
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?
Question Description
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?.
Solutions 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? in English & in Hindi are available as part of our courses for Class 11.
Download more important topics, notes, lectures and mock test series for Class 11 Exam by signing up for free.
Here you can find the meaning 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? defined & explained in the simplest way possible. Besides giving the explanation 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?, 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
ample number of questions to practice 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? tests, examples and also practice Class 11 tests.
|
|
Explore Courses for Class 11 exam
|
|
Suggested Free Tests
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
|
© EduRev
|
Education Revolution
|
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup