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