The beta-decay process,discovered around 1900 | vedclass

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(B) $1.$ In $\beta$-decay,the total energy $Q$ is shared between the proton,electron,and anti-neutrino: $Q = KE_p + KE_e + KE_{\bar{
u}}$. Since the

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$(B, C)$

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(B) $1.$ In $\beta$-decay,the total energy $Q$ is shared between the proton,electron,and anti-neutrino: $Q = KE_p + KE_e + KE_{\bar{
u}}$. Since the proton is very heavy,its recoil energy $KE_p$ is negligible. Thus,$Q \approx KE_e + KE_{\bar{
u}}$. The maximum energy of the anti-neutrino occurs when the electron's kinetic energy is zero,which is $KE_{\bar{
u}, \max} \approx Q = 0.8 	imes 10^6 \ eV$. Therefore,option $(C)$ is correct.$2.$ If the anti-neutrino has a non-zero mass $m_{\bar{
u}} = 3 \ eV/c^2$,the total energy $Q$ must also account for the rest mass energy of the anti-neutrino. The electron's kinetic energy $K$ is maximum when the anti-neutrino is at rest (its kinetic energy is zero). Thus,$K_{\max} = Q - m_{\bar{
u}}c^2$. Since $m_{\bar{
u}}c^2 = 3 \ eV$,$K_{\max} = 0.8 	imes 10^6 \ eV - 3 \ eV$. The electron's kinetic energy $K$ can range from $0$ (when the anti-neutrino carries away the maximum possible energy) to $K_{\max}$ (when the anti-neutrino is at rest). Thus,$0 \leq K < 0.8 	imes 10^6 \ eV$. Therefore,option $(D)$ is correct.

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