SSC
PHYSICS
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Question
[CLICK ON ANY CHOICE TO KNOW THE RIGHT ANSWER]
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Radium-226 undergoes alpha decay to become radon-222. The equation is:226 Ra 222 Rn + 4 He. 88 86 2Calculate the energy released if Ra-226 = 226.025406 u Rn-222 = 222.017574 u He = 4.002603 u
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7.81 x 10-11 J
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7.81 x 10-12 J
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7.81 x 10-13 J
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7.81 x 10-14 J
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Explanation:
Detailed explanation-1: -As a result, the atomic number of the parent atom is reduced by 2 and its mass number reduced by 4. Therefore, the nuclear equation describing the alpha decay of 22286Rn 86 222 R n is given by: 22286Rn→ 42He+21884Po 86 222 R n → 2 4 H e + 84 218 P o .
Detailed explanation-2: -Answer and Explanation: The balanced nuclear equation for alpha emission: 22388Ra→21986Rn+42
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