What is the energy (in MeV) released in the alpha decay of 231Pa?

3729.87MeV

Explanation:

²³¹₉₁Pa → ²²⁷₈₉Ac + ⁴₂He

Molar mass of of He = 4.0

Number of moles = 1

Na (Avogadro's number) = 6.023*10²³ atoms

Note : 1 mole = molarmass = Avogadro's number

Molar mass = 4.000 * 1 = 4g = 4*10^-3 kg

Mass = (number of moles * molar mass ) / Avogadro's number.

Mass = 4.0*10^-3 / 6.023*10^-23

Mass = 6.64 * 10^-27 kg

from Einstein's theory of relativity, energy E and mass m are interconvertible.

∇E = ∇mc²

E = (6.64*10^-27) * (3.0*10⁸)²

E = 5.98*10^-10J

1eV = 1.602*10-19J

XeV = 5.98*10-19J

X = (5.98*10^-10 * 1) / 1.602*10-19

X = 37328333958eV

X = 3729.87MeV

can be answered using momentum.

momentum = p

p1= mass1 * velocity1

p1 = p2

therefore

p1 = 6kg * 4m/s

p1 = 24kgm/s

p2=m2 * v2

24 kgm/s = 7kg * v2

v2 = 24 kgm/s / 7kg

v2= 3.42 m/s