Answer:
Speed of electron when their separation increased by a factor of 4.10 is 9.41 x 10⁵ m/s .
Explanation:
The electric potential energy is given by the relation :
![U = \frac{kq_{1}q_{2} }{r}](https://tex.z-dn.net/?f=U%20%3D%20%5Cfrac%7Bkq_%7B1%7Dq_%7B2%7D%20%20%7D%7Br%7D)
Here q₁ and q₂ are the two charge particles and r is the distance between them and k is electric constant.
In this case, there are two electrons which are separated by the distance 4.32 x 10⁻¹⁰ m.
Let e be the electron charge and r₁ be the distance between them. Then, the initial electric potential energy is :
![U_{1} = \frac{ke^{2} }{r_{1} }](https://tex.z-dn.net/?f=U_%7B1%7D%20%20%3D%20%5Cfrac%7Bke%5E%7B2%7D%20%20%20%7D%7Br_%7B1%7D%20%7D)
Now, the distance between the electrons increases by the factor of 4.10. Let r₂ be the new distance between them i.e. r₂ = 4.10 r₁.
Thus, the new electric potential energy is :
![U_{2} = \frac{ke^{2} }{r_{2} }=\frac{ke^{2} }{4.10r_{1} }](https://tex.z-dn.net/?f=U_%7B2%7D%20%20%3D%20%5Cfrac%7Bke%5E%7B2%7D%20%20%20%7D%7Br_%7B2%7D%20%7D%3D%5Cfrac%7Bke%5E%7B2%7D%20%20%20%7D%7B4.10r_%7B1%7D%20%7D)
Applying law of conservation of energy :
ΔU = ΔK
Here ΔU is change in electric potential energy and ΔK is change in kinetic energy.
( U₁ - U₂ ) = ( K₂ - K₁ )
Here K₂ and K₁ are initial and final kinetic energy of electron.
Since, the electron initially is at rest, so its initial kinetic energy is zero. Thus, the above equation becomes:
K₂ = U₁ - U₂
![\frac{1}{2}mv^{2}=\frac{ke^{2} }{r_{1} }- \frac{ke^{2} }{4.10r_{1} }](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7Dmv%5E%7B2%7D%3D%5Cfrac%7Bke%5E%7B2%7D%20%20%20%7D%7Br_%7B1%7D%20%7D-%20%5Cfrac%7Bke%5E%7B2%7D%20%20%20%7D%7B4.10r_%7B1%7D%20%7D)
Here m and v are the mass and final speed of electron respectively.
![v^{2}=\frac{2}{m} \frac{ke^{2} }{r_{1} }(1- \frac{1 }{4.10 })](https://tex.z-dn.net/?f=v%5E%7B2%7D%3D%5Cfrac%7B2%7D%7Bm%7D%20%5Cfrac%7Bke%5E%7B2%7D%20%20%20%7D%7Br_%7B1%7D%20%7D%281-%20%5Cfrac%7B1%20%20%7D%7B4.10%20%7D%29)
Substitute 9.1 x 10⁻³¹ kg for m, 9 x 10⁹ N m² C⁻² for k, 1.6 x 10⁻¹⁹ C for e and 4.32 x 10⁻¹⁰ m for r₁ in the above equation.
![v^{2}=\frac{2}{9.1\times10^{-31} } \frac{9\times10^{9}\times(1.6\times10^{-19})^{2} }{4.32\times10^{-10} }(1- \frac{1 }{4.10 })](https://tex.z-dn.net/?f=v%5E%7B2%7D%3D%5Cfrac%7B2%7D%7B9.1%5Ctimes10%5E%7B-31%7D%20%7D%20%5Cfrac%7B9%5Ctimes10%5E%7B9%7D%5Ctimes%281.6%5Ctimes10%5E%7B-19%7D%29%5E%7B2%7D%20%20%20%7D%7B4.32%5Ctimes10%5E%7B-10%7D%20%7D%281-%20%5Cfrac%7B1%20%20%7D%7B4.10%20%7D%29)
![v^{2}=8.86\times10^{11}](https://tex.z-dn.net/?f=v%5E%7B2%7D%3D8.86%5Ctimes10%5E%7B11%7D)
v = 9.41 x 10⁵ m/s