Question

An unstable atomic nucleus has a mass of 17.010-27kg, and starts out at rest. When it decays, it the original nucleus disintegrates into three particles. One particle has a mass of 5.0010-27kg and velocity of 6.00106 jm/s, and a second particle has a mass of 8.4010-27kg and velocity of 4.00106 im/s. Calculate (a) the velocity of the third particle and (b) the total increase in kinetic energy as a result of the decay. (Note that this kinetic energy actually comes from the mass of the original nucleus, according to E = ∆mc2, but only a tiny amount of mass is turned into energy, so we don’t have to worry about it for this problem.)

Answer 1

Part a)

`v = -(8.33\hat j + 9.33\hat i)* 10^6 m/s`

Part b)

`E = 4.4 * 10^(-13) J`

Step-by-step explanation:

As per momentum conservation we know that there is no external force on this system so initial and final momentum must be same

So we will have

`m_1v_1 + m_2v_2 + m_3v_3 = 0`

`(5 * 10^(-27))(6 * 10^6\hat j) + (8.4 * 10^(-27))(4 * 10^6\hat i) + (3.6 * 10^(-27)) v = 0`

`(30\hat j + 33.6\hat i)* 10^6 + 3.6 v = 0`

`v = -(8.33\hat j + 9.33\hat i)* 10^6 m/s`

Part b)

By equation of kinetic energy we have

`E = (1)/(2)m_1v_1^2 + (1)/(2)m_2v_2^2 + (1)/(2)m_3v_3^2`

`E = (1)/(2)(5 * 10^(-27))(6* 10^6)^2 + (1)/(2)(8.4 * 10^(-27))(4 * 10^6)^2 + (1)/(2)(3.6 * 10^(-27))(8.33^2 + 9.33^2) * 10^(12)`

`E = 9* 10^(-14) + 6.72 * 10^(-14) + 2.82* 10^(-13)`

`E = 4.4 * 10^(-13) J`