e*100,000 is the kinetic energy.
e is the electron charge,1.6*10^-19 C
Set 100,000*e equal to (1/2)mV^2 and solve for V
m is the electron mass.
if electron are caused to fall through a potential difference of 100000 voltage , determine their final speed if they were initially at rest.
6 answers
6646
p.d.=100000V
therefore, workdone , W=q×p.d.
W=1.6×10^-19 × 100000
=1.6×10^-14 Joules
since work done = force × displacement
F×s=1.6×10^-14
m×a×s=1.6×10^-14
9.1×10^-31×a×s=1.6×10^-14
a×s=(1.6×10^-14)÷(9.1×10^-31)
a×s=1.75×10^17
v^2/2=1.75×10^17
v=5.09×10^8m/s
therefore, workdone , W=q×p.d.
W=1.6×10^-19 × 100000
=1.6×10^-14 Joules
since work done = force × displacement
F×s=1.6×10^-14
m×a×s=1.6×10^-14
9.1×10^-31×a×s=1.6×10^-14
a×s=(1.6×10^-14)÷(9.1×10^-31)
a×s=1.75×10^17
v^2/2=1.75×10^17
v=5.09×10^8m/s
Recall that one electron-volt is the energy gained by an electron in moving through a potential difference of 1 V, so
1 eV = 1.6 * 10^-19 Joules
1,000,000 V = 1.6 ^ 10^-19 * 10^6 Joules = 1.6 * 10^-13 J
So this must be the kinetic energy gained be the electron.
KE = (1/2) m v^2
From reference tables
mass of electron = 9.11 * 10^-31 kg
So
1.6 * 10^-13 = (1/2) * 9.11 * 10^-31 * v^2
v^2 = 2 * 1.6 * 10^-13 / 9.11 * 10^-31 = 0.35 * 10^18
v = 0.59 * 10^9 m/s
1 eV = 1.6 * 10^-19 Joules
1,000,000 V = 1.6 ^ 10^-19 * 10^6 Joules = 1.6 * 10^-13 J
So this must be the kinetic energy gained be the electron.
KE = (1/2) m v^2
From reference tables
mass of electron = 9.11 * 10^-31 kg
So
1.6 * 10^-13 = (1/2) * 9.11 * 10^-31 * v^2
v^2 = 2 * 1.6 * 10^-13 / 9.11 * 10^-31 = 0.35 * 10^18
v = 0.59 * 10^9 m/s
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By deriving a formula ,from work energy theorem,we get:- V=√{2×e×v×}/m
Where,m=mass of electron
e=charge on an electron
v=potential difference
V=final velocity of an electron
Where,m=mass of electron
e=charge on an electron
v=potential difference
V=final velocity of an electron