Question

A thorium atom of mass 232.038 u decays by the emission of an alpha particle to a radium atom of mass 228.031 u. If the alpha particle has a mass of 4.0026 u, how much energy in J is released in the process? (Answer is 6.56 x 10^-13 J)

Answer 1

Energy released in above reaction will be given by:

Q = dm*c^2

Q = (m_Th - m_Ra - m_He)*c^2

m_Th = mass of Thorium = 232.038 u

m_Ra = mass of Radium = 228.031 u

m_He = mass of Helium/alpha particle = 4.0026 u

Mass defect will be

dm = 232.038 u - 228.031 u - 4.0026 u = 0.0044 u

Also we know that

1 u = 931.5 MeV/c^2
Using these values:

Q = (0.0044)*(931.5 MeV/c^2)*c^2

Q = 4.0986 MeV

Now Change above energy from MeV to J

1 MeV = 1*10^6 eV = 1*10^6*1.6*10^-19 J

So,

Q = 4.0986*1*10^6*1.6*10^-19 J

Q = 6.56*10^-13 J

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