An electron with an initial speed of 660,000 m/s slows to a stop due to an electric field. The electron must have moved from a region of higher potential to a region of lower potential because, as a negative charge, it slows as it travels. The potential difference that stopped the electron was 1.24 V. The initial kinetic energy of the electron was 1.24 eV.

1. tem 9 Constants Part A An electron with an initial speed of 660,000 m/s is brought to rest by an
electric field. Did the electron move into a region of higher potential or lower potential? Because
the electron is a negative charge and it slows down as it travels, it must be moving from a region
of lower potential to a region of higher potential. OBecause the electron is a positive charge and
it accelerates as it travels, it must be moving from a region of higher potential to a region of
lower potential Because the electron is a negative charge and it slows down as it travels, it must
be moving from a region of higher potential to a region of lower potential Because the electron is
a positive charge and it accelerates as it travels, it must be moving from a region of lower
potential to a region of higher potential Submit Request Answer Part B What was the potential
difference that stopped the electron? Express your answer to two significant figures and include
the appropriate units. ??_ | Value II Units Submit Request Answer Part C What was the initial
kinetic energy of the electron, in electron volts? Express your answer using two significant
figures Ki- eV Submit Request Answer
Solution
initial speed, u=66*10^4 m/sec
A)
the electron move into a region of lower potential
==> because the electron is a negative charge and it slows down as it travels, it must be moving
from a region of higher potential region to lower potential region
B)
K2-K1=q*dv
0-1/2*m*u^2 =q*dV

2. ==> 1/2*9.1*10^-31*(6.6*10^5)^2=1.6*10^-19*dV
==> dV=1.24 v
potential difference, dV=1.24 v
C)
K.E=1/2*m*v^2
=1/2*9.1*10^-31*(6.6*10^5)^2
=1.982*10^-19 J
=1.24 ev