Question

You may want to review (Pages 1132-1134) Submitted Answers ANSWER 1: Deduction:-3% Part A 59 nm What is the minimum uncertainty in position, in nm, of an electron whose velocity is known to be between 2.34x105 m/s and 3.32 105 m/s? Express your answer using two significant figures. You may have used the reduced Planck constant in the uncertainty principle. Note that in statements of the uncertainty principle the right side could have different meanings (just h or it could contain various factors of PI), but we have h/2 59 nm ANSWER 2 Deduction:-3% 7.4 nm SubmitD Previous Answers Request Answer You may have used the reduced Planck constant in the uncertainty principle. Note th Incorrect; Try Again;4 attempts remaining of the uncertainty principle at in statements the right side could have different meanings just h or it could contain various factors of PI), but we have h/2. < Return to Assignment Provide Feedback

Answer 1

Delta x Delta p greaterthanorequalto h/2 (According to uncertainty principle) Delta x times m Delta v = h/2 (For minimum uncertainty in position). Delta x = h/2 m Delta v Delta v = (3.32 - 2.34) times 10^5 = 0.98 times 10^5 Delta x = 1.05 times 10^-34 - 5/2 times 9.11 times 10^-31 times 0.98 = 1.05 times 10^-34 - 5 + 31/2 times 9.11 times 0.98 = 0.05880508 times 10^-8 \r\n Delta x Delta p greaterthanorequalto h/2 Delta x times m times Delta v = h/2 (For minimum uncertainty in position) Delta x = h/2 m Delta v = 6.634 times 10^-34/2 times 9.11 times 10^-31 times 0.98 times 10^5 Delta v = (3.32 - 2.34) times 10^5 = 0.98 times 10^5 Delta x = 6.634 times 10^-34 + 31 - 5/2 times 9.11 times 0.98 = 0.3715361 times 10^-8 Delta x = 3.715361 nm