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Question

(B) For a nuclear decay to be spontaneous,the $Q$-value of the reaction must be positive,which implies that the mass of the parent nucleus must be gre

Vedclass · Accepted Answer

$M(A, Z) > M(A, Z+1)$

Vedclass · Answer

(B) For a nuclear decay to be spontaneous,the $Q$-value of the reaction must be positive,which implies that the mass of the parent nucleus must be greater than the sum of the masses of the daughter nucleus and the emitted particles.The decay process is: $X_Z^A ightarrow Y_{Z+1}^A + e^{-} + \bar{
u}_e$.Let $M_n(A, Z)$ be the mass of the nucleus. The condition is: $M_n(A, Z) > M_n(A, Z+1) + m_e$ (neglecting neutrino mass).Since $M(A, Z)$ represents the mass of the neutral atom,we have $M_n(A, Z) = M(A, Z) - Z m_e$.Substituting this into the condition:$(M(A, Z) - Z m_e) > (M(A, Z+1) - (Z+1) m_e) + m_e$.Simplifying the right side:$M(A, Z) - Z m_e > M(A, Z+1) - Z m_e - m_e + m_e$.$M(A, Z) - Z m_e > M(A, Z+1) - Z m_e$.Therefore,$M(A, Z) > M(A, Z+1)$.

Similar Questions

Which one of the following is a possible nuclear reaction?

$\beta$-decay means emission of electron from

The nucleus $_{10}^{23} Ne$ decays by $\beta^{-}$ emission. Write down the $\beta^{-}$-decay equation and determine the maximum kinetic energy of the electrons emitted. Given that:
$m(_{10}^{23} Ne) = 22.994466 \; u$
$m(_{11}^{23} Na) = 22.989770 \; u$

Consider the following two statements:
$A.$ The energy spectrum of $\alpha-$ particles emitted in radioactive decay is discrete.
$B.$ The energy spectrum of $\beta-$ particles emitted in radioactive decay is continuous.

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