Question

An electron is moving in the field of a helium nucleus (Q = +2e).

1,) What is the change in the electrons potential energy when it moves from a circular orbit of radius 3

Answer 1

1)

U= F*d= [Kq₁q₂/d²]* d= Kq₁q₂/d

so from the above equation, electron potential energy, U is inversely proportional to d

U ~1/d

U₁= 9*10⁹* 1.6*10^-19* 2* 1.6*10^-19/3*10^-10 = 15.36*10^-19J

U₂ =  9*10⁹* 1.6*10^-19* 2* 1.6*10^-19/2*10^-10 = 23.04*10^-19J

change in potential energy, delta U= U₂ -U₁= 7.68*10^-19J

2) Using the centripetal force, we get the expression for velocity as,

m_ev²/r= kq₁q₂/r² [ m_e- mass of electron, q₁ charge of electron and q₂-charge of helium]

v²= kq₁q₂ / mr

v= [  kq₁q₂ / mr ] ^1/2

For r₁= 3*10^-10m, v₁= 1.3*10⁶ m/s

For r₂= 2*10^-10m, v₂= 1.6*10⁶ m/s

KE₁= 1/2m_ev₁²= 7.7*10^-19J

KE₂= 1/2m_ev₂²= 11.7*10^-19J

Change in KE= KE₂-KE₁= 4*10^-19J

3)

The law of conservation of energy states that the total amount of energy in a system remains constant although energy within the system can be changed from one form to another or transferred from one object to another. Energy cannot be created or destroyed, but it can be transformed.