Question

The Americium nucleus, 241 95 Am, decays to a Neptunium nucleus, 237 93 Np, by emitting an alpha particle of mass 4.00260 u and kinetic energy 5.5 MeV. Estimate the mass of the Neptunium nucleus, ignoring its recoil, given that the Americium mass is 241.05682 u

Answer 1

Answer:

The mass of Neptunium is 237.054 u.

Explanation:

Given that,

Mass of Americium = 241.05682 u

Mass of alpha particle = 4.00260 u

The equation is,

$_{95}^{241}Am\rightarrow _{93}^{237}NP+\alpha\ particle+5.5 MeV$

Let the mass of Neptunium is m.

Since the mass remain same.

We need to calculate the mass of Neptunium

Using formula of mass

Mass of Neptunium =  Mass of Americium -Mass of alpha particle

Put the value into the formula

$m=241.05682 -4.00260$

$m=237.054\ u$

Hence, The mass of Neptunium is 237.054 u.