How many joules are released when one atom of americium 241, the isotope used in ionization-type smoke detectors, undergoes apha emission?

When Americium (Am-241) undergoes alpha decay (He-4) it forms neptunium (Np-237) based on the following pathway:

²⁴¹Am₉₅ → ²³⁷Np₉₃ + ⁴He₂

The energy released in given as:

ΔE = Δmc²

where Δm = mass of products - mass of reactants

= [m (Np-237) + m (He-4) ] - [m (Am-241) ]

= 237.0482+4.0015-241.0568 = - 0.0073 g/mol = - 7.3 * 10⁻⁶ kg/mol

ΔE = - 7.3*10⁻⁶ kg/mol * (3*10⁸ m/s) ² = - 5.84*10¹¹ J/mol