What is the escape speed of an electron launched from the surface of a 1.1-cm-diameter glass sphere that has been charged to 8.0 nc ?

Respuesta :

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At surface,
v = kq/r

And potential energy of an electron is given by,
PE = -ev = -ekq/r

At escape velocity,
PE + KE = 0.
Therefore,
1/2mv^2 - ekq/r =0
1/2mv^2 = ekq/r
v = Sqrt [2ekq/mr], where v = escape velocity, e = 1.6*10^-19 C, k = 8.99*10^9 Nm^2/C^2, m = 9.11*10^-31 kg, r = 1.1*10^-2 m, q = 8*10^-9 C

Substituting;
v = Sqrt [(2*1.6*19^-19*8.99*10^9*8*10^-9)/(9.11*10^-31*1.1*10^-2)] = 47949357.23 m/s ≈ 4.795 *10^7 m/s

It is the velocity required to get out from that partticular orbit in which the electron is revolving The escape velocity of the electron will be [tex]\rm { 4.795 \times 10^7 m/sec}[/tex]

What is the escape velocity of an electron?

It is the velocity required to get out from that partticular orbit in which the electron is revolving. For a given orbit value of escape velocity will be different.

At escape velocity, the sum of kinetic and potential energy is equal to zero.

KE + PE =0

[tex]KE = \frac{1}{2}mv^{2}[/tex]

[tex]PE = eV =-e\frac{Kq}{r}[/tex]

KE + PE =0

KE = -PE

[tex]\frac{1}{2}mv^{2}=e\frac{Kq}{r}[/tex]

[tex]V^{2} = \sqrt{\frac{2Keq}{rm} }[/tex]

[tex]\rm {V= \sqrt{\frac{2\times8.99\times(10)^{-9}\times1.6\times10^{-19}\times8\times10^{-9}}{1.1\times\times10^{-2}\times9.11\times10^{-31}} }}[/tex]

[tex]\rm {V = 4.795 \times 10^7 m/sec}[/tex]

the escape velocity of an electron will be [tex]\rm { 4.795 \times 10^7 m/sec}[/tex].

To learn more about the escape velocity refer to the link ;

https://brainly.com/question/9757724