Answer:
time will elapse before it return to its staring point is 23.6 ns
Explanation:
given data
speed u = 2.45 ×
m/s
uniform electric field E = 1.18 ×
N/C
to find out
How much time will elapse before it returns to its starting point
solution
we find acceleration first by electrostatic force that is
F = Eq
here
F = ma by newton law
so
ma = Eq
here m is mass , a is acceleration and E is uniform electric field and q is charge of electron
so
put here all value
9.11 ×
kg ×a = 1.18 ×
× 1.602 ×
a = 20.75 ×
m/s²
so acceleration is 20.75 ×
m/s²
and
time required by electron before come rest is
use equation of motion
v = u + at
here v is zero and u is speed given and t is time so put all value
2.45 ×
= 0 + 20.75 ×
(t)
t = 11.80 ×
s
so time will elapse before it return to its staring point is
time = 2t
time = 2 ×11.80 ×
time is 23.6 ×
s
time will elapse before it return to its staring point is 23.6 ns
C..............................
Answer:
h f = Wf + K
where the total energy available is h f, Wf is the work function or the work needed to remove the electron and K is the kinetic energy of the removed electron
If K = zero then hf = Wf
Wf = h f = h c / λ or
λ = h c / Wf = 6.63E-34 * 3.0E8 / (3.7 * 1.6E-19)
λ = 6.63 * 3 / (3.7 * 1.6) E-7 = 3.36E-7
This would be 3360 angstroms or 336 millimicrons
Visible light = 400-700 millimicrons
Your answer will be Radio Waves .
That seems to be the only to make sense. Hope that helps u
I don’t think I’m right but I want to say 500 m/s