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2 years ago

6

The number of significant figures on the measurement 0.050010 kg id

1 answer:

Alex17521 [72]2 years ago

6 0

Its been some time so i might be wrong but i think the answer is 3 either or 2

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Arlecino [84]

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The radius is $r = 4.434 *10^{-5} \ m$

Explanation:

From the question we are told that

The magnetic field is $B = 90 mT = 90*10^{-3} \ T$

The electron kinetic energy is $KE = 1.4 eV = 1.4 * (1.60*10^{-19}) =2.24*10^{-19} \ J$

Generally for the collision to occur the centripetal force of the electron in it orbit is equal to the magnetic force applied

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$\frac{mv^2}{r} = qvB$

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$v = \sqrt{\frac{2 * KE}{m} }$

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$r = \frac{m}{qB} * \sqrt{\frac{2 * KE}{m} }$

apply indices

$r = \frac{\sqrt{2 * KE * m} }{qB}$

substituting values

$r = \frac{\sqrt{2 * 2.24*10^{-19}* 9.1 *10^{-31}} }{ 1.60 *10^{-19}* 90*10^{-3}}$

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