Respuesta :
Answer:
The speed of electron is [tex]1.5\times10^{7}\ m/s[/tex]
Explanation:
Given that,
Electric field [tex]E=1.50\times10^{3}\ N/C[/tex]
Distance = -0.0200
The electron's speed has fallen by half when it reaches x = 0.190 m.
Potential energy [tex]P.E=5.04\times10^{-17}\ J[/tex]
Change in potential energy [tex]\Delta P.E=-9.60\times10^{-17}\ J[/tex] as it goes x = 0.190 m to x = -0.210 m
We need to calculate the work done by the electric field
Using formula of work done
[tex]W=-eE\Delta x[/tex]
Put the value into the formula
[tex]W=-1.6\times10^{-19}\times1.50\times10^{3}\times(0.190-(-0.0200))[/tex]
[tex]W=-5.04\times10^{-17}\ J[/tex]
We need to calculate the initial velocity
Using change in kinetic energy,
[tex]\Delta K.E = \dfrac{1}{2}m(\dfrac{v}{2})^2+\dfrac{1}{2}mv^2[/tex]
[tex]\Delta K.E=\dfrac{-3mv^2}{8}[/tex]
Now, using work energy theorem
[tex]\Delta K.E=W[/tex]
[tex]\Delta K.E=\Delta U[/tex]
So, [tex]\Delta U=W[/tex]
Put the value in the equation
[tex]\dfrac{-3mv^2}{8}=-5.04\times10^{-17}[/tex]
[tex]v^2=\dfrac{8\times(5.04\times10^{-17})}{3m}[/tex]
Put the value of m
[tex]v=\sqrt{\dfrac{8\times(5.04\times10^{-17})}{3\times9.1\times10^{-31}}}[/tex]
[tex]v=1.21\times10^{7}\ m/s[/tex]
We need to calculate the change in potential energy
Using given potential energy
[tex]\Delta U=-9.60\times10^{-17}-(-5.04\times10^{-17})[/tex]
[tex]\Delta U=-4.56\times10^{-17}\ J[/tex]
We need to calculate the speed of electron
Using change in energy
[tex]\Delta U=-W=-\Delta K.E[/tex]
[tex]\Delta K.E=\Delta U[/tex]
[tex]\dfrac{1}{2}m(v_{f}^2-v_{i}^2)=4.56\times10^{-17}[/tex]
Put the value into the formula
[tex]v_{f}=\sqrt{\dfrac{2\times4.56\times10^{-17}}{9.1\times10^{-31}}+(1.21\times10^{7})^2}[/tex]
[tex]v_{f}=1.5\times10^{7}\ m/s[/tex]
Hence, The speed of electron is [tex]1.5\times10^{7}\ m/s[/tex]
