## Test 3, Question #1

$\Delta p \Delta x\geq \frac{h}{4\pi }$

Hena Sihota 1L
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### Test 3, Question #1

How do you solve this problem?
A professor is walking towards the class. An electron in his body has a speed of 1.5 +/- 0.4 m/s. What is the uncertainty in the position of the electron?
I got deltax= (1.05457x10^(-34))/(2x(9.1095x10^(-31))x(2.3))= 2.5x10^(-5) m, but the correct answer is 7.35x10^(-5) m

Ishan Saha 1L
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Joined: Fri Sep 29, 2017 7:03 am

### Re: Test 3, Question #1

Hi! So because the uncertainty in velocity is 1.5 +/- 4m/s, the delta v is 0.8 m/s. To find delta p, you multiply delta v by the mass of an electron (9.11 x10^-31 kg). Then to find delta x, you do h/(4pi)/delta p which numerically is (6.626 x 10 ^-34)/(4 x pi)/(0.8 x (9.11 x 10-31)) and you get the correct answer.

Hena Sihota 1L
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### Re: Test 3, Question #1

So I don't use the 1.5 anywhere?

Dylan Mai 1D
Posts: 70
Joined: Sat Jul 22, 2017 3:00 am

### Re: Test 3, Question #1

the deltav is just (1.5+0.4)-(1.5-0.4). You could use the 1.5 in this calculation, or disregard it and just double 0.4 for your delta v