Question

An electron is to be accelerated from a velocity of 2.50×10⁶ m/s to a velocity of 9.00×10⁶ m/s . Through what potential difference must the electron pass to accomplish this?

Through what potential difference must the electron pass if it is to be slowed from 9.00×10⁶ m/s to a halt?

Answer 1

Solution)

Potential Difference, V= m/2q (vt^2 -vi^2)

V= (9.1*10^-31)/2(-1.6*10^-19)*((9*10^6)^2-(2.5*10^6)^2)

V=-212.57 V

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Now, v= 0

So, V= (9.1*10^-31)/2(-1.6*10^-19)*(0^2-(9*10^6)^2)

V= 230.34 V

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