A common example of alpha decay is 238 92 U → 234 90 Th + 2 He 4 + Q Given: 238 92 U = 238 . 05060   u 234 90 Th = 234 . 04360   u 4 2 He = 4 . 00260   u and 1 u = 931 . 5 MeV c 2 The energy released Q during the alpha

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A common example of alpha decay is 238 92 U &#8594; 234 90 Th + 2 He 4 + Q Given: 238 92 U = 238 . 05060 &#160; u 234 90 Th = 234 . 04360 &#160; u 4 2 He = 4 . 00260 &#160; u and 1 u = 931 . 5 MeV c 2 The energy released Q during the alpha decay of 238 92 U is _____ MeV .

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The formula to calculate the Q-value of the given nuclear reaction can be written as Q = m U - m Th - m He c 2 &#160; &#160; &#160; . . . 1 Substitute the values of the known parameters into equation (1) to calculate the required Q-value for the reaction. Q = 238 . 05060 &#160; u - 234 . 04360 &#160; u - 4 . 00260 &#160; u c 2 = 0 . 0044 c 2 &#215; 931 . 5 &#160; MeV / c 2 &#8776; &#160; 4 &#160; MeV

A common example of alpha decay is 238 92 U → 234 90 Th + 2 He 4 + Q Given: 238 92 U = 238 . 05060   u 234 90 Th = 234 . 04360   u 4 2 He = 4 . 00260   u and 1 u = 931 . 5 MeV c 2 The energy released Q during the alpha decay of 238 92 U is _____ MeV .

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A common example of alpha decay is 238 92 U &#8594; 234 90 Th + 2 He 4 + Q Given: 238 92 U = 238 . 05060 &#160; u 234 90 Th = 234 . 04360 &#160; u 4 2 He = 4 . 00260 &#160; u and 1 u = 931 . 5 MeV c 2 The energy released Q during the alpha decay of 238 92 U is _____ MeV .

Correct answer

Step-by-step solution

More from Nuclear Physics

A common example of alpha decay is 238 92 U → 234 90 Th + 2 He 4 + Q Given: 238 92 U = 238 . 05060 u 234 90 Th = 234 . 04360 u 4 2 He = 4 . 00260 u and 1 u = 931 . 5 MeV c 2 The energy released Q during the alpha decay of 238 92 U is _____ MeV .