Neon-23 undergoes beta decay in the following way: 10Ne23→11Na23+ −1e0+v→ Atomic masses of 10Ne23 and N 11 a23 are 22.9945u and 22.9898u respectively. The minimum KE of β particle is 1.1x MeV. (1u=931.5 Me V/c2), then x=

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Neon-23 undergoes beta decay in the following way: $_{10} N e^{23} \to_{11} N a^{23} +_{- 1} e^{0} + \vec{v}$ Atomic masses of $_{10} N e^{23}$ and ${{}_{11}N a}^{23}$ are 22.9945u and 22.9898u respectively. The minimum KE of $β$ particle is 1.1x MeV. (1u=931.5 Me $V / c^{2}$ ), then x=

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Detailed Solution

The Q value of the reaction is:
$Q = Δ m \times C^{2} Δ m = 22.9945 - 22.9898 = 0.0047 ∴ Δ Q = 0.0047 - 931.5 M e V = 4.4 M e V$
This released energy is shared between electron (beta particle) and antineutrino. The minimum KE of $β -$ particle is 0 and maximum KE of $β -$ particle is 4.4 MeV.