Chapter 26, Problem 39P

To determine

The mass of the neptunium nucleus.

Answer to Problem 39P

  $237.04832u$ .

Explanation of Solution

Given:

Kinetic energy, $KE=5.5MeV$

Mass of Americium nucleus, $m_{Am}=241.05682u$

Mass of alpha, $m_{\alpha }=4.00260u$

Formula used:

If the recoil of the neptunium nucleus is ignored, then the increase in kinetic energy is the kinetic energy of the alpha particle,

  $K{E}_{\alpha }=\left[m_{Am}-\left(m_{Np}+m_{\alpha }\right)\right]c^2$

Where,

  $m_{Am}$ is Mass of Americium nucleus

  $m_{Np}$ is Mass of neptunium nucleus

  $m_{\alpha }$ is Mass of alpha particle

c is speed of light, $3.00\times {10}^{8}m/s$

Calculation:

Substitute all the given values

  $5.5MeV=\left[241.05682u-\left(m_{Np}+4.00260u\right)\right]c^2\left(\frac{931.5MeV}{u{c}^2}\right)$

  $m_{Np}=237.04832u$

Conclusion:

The required value is $237.04832u$