Physics question 9 hw 21

Physics
Tutor: None Selected Time limit: 1 Day

A proton (mp = 1.673 × 10–27 kg) and an electron (me = 9.109 × 10–31 kg) are confined such that the x position of each is known to within 1.50 × 10–10 m. What is the ratio of the minimum uncertainty in the x component of the velocity of the electron to that of the proton, Δve/Δvp?

Jul 15th, 2015

Thank you for the opportunity to help you with your question!

uncertainty principle gives

 delta x*deltap>h/2pi

but deltap=deltam*deltav

delta (ve)=h/2pi*delta(m)*delta(x)

               

so delta(ve)/delta (vp) =delta (mp)/delta (me)

                                    =1.673*10^-27/9.109*10^-31

                                   =0.18 *10^4


Please let me know if you need any clarification. I'm always happy to answer your questions.
Jul 15th, 2015

answer is 1837

Jul 15th, 2015

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