Physics question 11 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, Δvevp?

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

                                 




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Jul 15th, 2015

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