The x coordinate of an electron is measured with an uncertainty of 0.200mm . Wha

Question

The x coordinate of an electron is measured with an uncertainty of 0.200mm . What is vx, the x component of the electron's velocity, if the minimum percentage uncertainty in a simultaneous measurement of vx is 1.00% ? Use the following expression for the uncertainty principle:

?x?px??,

where ?x is the uncertainty in the x coordinate of a particle, ?px is the particle's uncertainty in the x component of momentum, and ?=h2?, where h is Planck's constant.

Tried prevouis asked solutions none seem to be correct. Will rate ASAP!!! Thank you!!!

Explanation / Answer

delta p_x = h/(2pi*delta x)

=(6.626*10^-34)/(2*pi*0.0002)

=5.272*10^-31

delta p_x=m* delta v_x

mass of electron = 9.1*10^-31 kg

delta v_x = delta p_x/m

=(5.272*10^-31)/(9.1*10^-31)

=0.57934

delta v_x = 1% of v_x = 0.01 v_x

So,

v_x = delta v_x /0.01 = 0.57934/0.01

=57.934 m/second

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