00:01
For the given problem, radium 2 -26 decays into alpha plus radon, radon 2 -2, where alpha has got a charge of q is equal to plus 2e, and the radon has got 86 proton, and its charge will be then q of redone charge will be plus 86, e.
00:35
We are also given the energy of the alpha particles where it is infinitely far from the nucleus.
00:42
This means its potential energy is a, let's call it a u2, which is zero, and this energy is the kinetic energy k2 that is 4 .79 mega electron volts.
01:01
After the decay, the alpha particles move away from the nucleus of the radon.
01:08
So we have travel and the total mechanical energy during the travel is a conservative.
01:17
We can apply the law of conservation of energy that is a k1 plus u1 is equal to k2 plus u2, where k1 and k2 are the kind, sorry, k1 and u2 are the kinetic and potential energy of the alpha radon combination before decay.
01:41
So our target now is to calculate u2, where before the decay, the kinetic energy, k1, is equal to zero.
01:52
From the equation, then we can write 0 plus u1 is equal to k2, plus zero...