Question

Find the energy released in the alpha decay⁹²²³⁸U → ⁹⁰²³⁴Th+²4He

Answer 1

The energy released in the alpha decay⁹²²³⁸U → ⁹⁰²³⁴Th+²4He is 4.25MeV.

Solution:

Q =(M U −M TH −M He )$c^{2}$

  =(238.05079−234.04363−4.00260)u×$c^{2}$

   =(0.00456u)×c 2

  =0.00456×( 931−MeV/$c^{2}$).$c^{2}$ = 4.25MeV

Yes, the decay is spontaneous (since Q is positive).

To learn more about alpha decay, refer

brainly.com/question/1898040

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