Question

The deuterium nucleus starts out with a kinetic energy of 1.24 × 10-13 joules, and the proton starts out with a kinetic energy of 2.47 × 10-13 joules. The radius of a proton is 0.9 × 10-15 m; assume that if the particles touch, the distance between their centers will be twice that. What will be the total kinetic energy of both particles an instant before they touch?

Answer 1

Answer:

The total kinetic energy of both particles is $2.43\times10^{-13}$

Explanation:

Given that,

Kinetic energy of nucleus$K.E= 1.24\times10^{-13}\ J$

Kinetic energy of proton $K.E= 2.47\times10^{-13}\ J$

Radius of proton $r= 0.9\times10^{-15}\ m$

We need to calculate the final potential energy

Using formula of final potential energy

$U=\dfrac{kq^2}{r}$

Put the value into the formula

$U_{f}=\dfrac{9\times10^{9}\times(1.6\times10^{-19})^2}{2\times0.9\times10^{-15}}$

$U_{f}=1.28\times10^{-13}\ J$

We need to calculate the initial energy of both the particles

Using formula of energy

$E_{i}=(K.E_{n}+K.E_{p})+U_{i}$

$E_{i}=1.24\times10^{-13}+2.47\times10^{-13}+0$

$E_{i}=3.71\times10^{-13}\ J$

We need to calculate the total kinetic energy of both particles

Using conservation of energy

$E_{i}=E_{f}$

$E_{i}=K.E_{f}+U_{f}$

$3.71\times10^{-13}=K.E_{f}+1.28\times10^{-13}$

$K.E_{f}=3.71\times10^{-13}-1.28\times10^{-13}$

$K.E_{f}=2.43\times10^{-13}$

Hence, The total kinetic energy of both particles is $2.43\times10^{-13}$