an electron is emmited with a velocity of 5*10^6 m/s.it is accelerated by an electric field in the direction of initial velocity at 3*10^14 m/s. if its final velocity is 7*10^6 m/s. calculate the time taken by the electron to attain the final velocity and distance covered by it.
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Explanation:
Initial velocity of electron, u=5\times 10^6\ m/su=5×10
6
m/s
Final velocity of electron, v=7\times 10^6\ m/sv=7×10
6
m/s
Acceleration of electrons, a=3\times 10^{14}\ m/s^2a=3×10
14
m/s
2
We know from the third equation of motion :
v^2-u^2=2\ a\ sv
2
−u
2
=2 a s
where,
s is the displacement of the electron.
(7\times 10^6\ m/s)^2-(5\times 10^6\ m/s)^2=2\times 3\times 10^{14}\ m/s^2\times s(7×10
6
m/s)
2
−(5×10
6
m/s)
2
=2×3×10
14
m/s
2
×s
On solving,
s=4\times 10^{-2}\ ms=4×10
−2
m
or
s=0.04\ ms=0.04 m
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