Respuesta :
Answer:
electron moving at 91.82 m/s
Explanation:
given data
distance d1 = 6 cm
distance d2 = 3 cm
to find out
how fast will the electron be moving
solution
we know potential energy formula that is
potential energy = [tex]k\frac{q1q2}{r}[/tex]
here k is 9× [tex]10^{9}[/tex] N-m²/C²
m mass of electron is 9.11 × [tex]10^{-31}[/tex] kg
and q = 1.60 × [tex]10^{-19}[/tex] C
we consider here initial potential energy = U1
U1 = [tex]9*10^{9}\frac{1.60*10^{-19}*1.60*10^{-19}}{0.06^2}[/tex]
U1 = 3.84 × [tex]10^{-27}[/tex] J
and final potential energy = U2
U2 = [tex]9*10^{9}\frac{1.60*10^{-19}*1.60*10^{-19}}{0.03}[/tex]
U2 = 7.68 × [tex]10^{-27}[/tex] J
and speed of electron = v
so we will apply here conservation of energy
0.5×m×v² = U2 - U1 ................1
so
0.5×9.11× [tex]10^{-31}[/tex] ×v² = 3.84 × [tex]10^{-27}[/tex]
v = 91.82 m/s
so electron moving at 91.82 m/s
Answer:
The speed of the electron is 91.86 m/s.
Explanation:
Given that,
Distance of electron from proton = 6.00 cm
Distance of proton = 3.00 cm
We need to calculate the initial potential energy
[tex]U_{1}=\dfrac{kq_{1}q_{2}}{r}[/tex]
Put the value into the formula
[tex]U_{i}=\dfrac{9\times10^{9}\times(1.6\times10^{-19})^2}{6.00\times10^{-2}}[/tex]
[tex]U_{i}=3.84\times10^{-27}\ J[/tex]
The final potential energy
[tex]U_{f}=\dfrac{9\times10^{9}\times(1.6\times10^{-19})^2}{3.00\times10^{-2}}[/tex]
[tex]U_{f}=7.68\times10^{-27}\ J[/tex]
We need to calculate the speed of the electron
Using conservation of energy
[tex]\dfrac{1}{2}mv^2=U_{f}-U_{i}[/tex]
Put the value into the formula
[tex]\dfrac{1}{2}\times9.1\times10^{-31}\times v^2=7.68\times10^{-27}-3.84\times10^{-27}[/tex]
[tex]v=\sqrt{\dfrac{2\times3.84\times10^{-27}}{9.1\times10^{-31}}}[/tex]
[tex]v=91.86\ m/s[/tex]
Hence, The speed of the electron is 91.86 m/s.
