Question

The average distance an electron travels between collisions is 2.0 μm . What acceleration must an electron have to gain 2.0×10−18 J of kinetic energy in this distance? Express your answer to two significant figures and include the appropriate units. a = nothing nothing

Answer 1

`1.1* 10^(18) m/s^(2)`

Step-by-step explanation:

Work=Kinetic energy

Fd=KE but we know that F=ma hence

ma d= KE

`a=\frac {KE}{md}`

Here a is acceleration, d is distance, KE is kinetic energy, m is mass

Taking m as
`9.11* 10^(-31) Kg`, KE as
`2* 10^(-18)` then d as
`2* 10^(-6)` then

`a=\frac {2* 10^(-18)}{ 9.11* 10^(-31)* 2* 10^(-6)}=1.1* 10^(18) m/s^(2)`