Potassium-40 can decay in three modes. It can decay via β^ -...

Potassium-40 can decay in three modes. It can decay via

\beta^{-}

-emission,

\beta^{+}

-emission or electron capture. Determine the equations showing the end products for these three modes respectively.

Options

A $\beta^{-}: ^{40}_{19}\text{K} \to ^{40}_{20}\text{Ca} + e^{+} + \nu$
$\beta^{+}: ^{40}_{19}\text{K} \to ^{40}_{18}\text{Ar} + e^{-} + \bar{\nu}$
Electron capture: ^{40}_{19}\text{K} + e^{+} \to ^{40}_{18}\text{Ar} + \nu
B $\beta^{-}: ^{40}_{19}\text{K} \to ^{40}_{20}\text{Ca} + e^{-} + \bar{\nu}$
$\beta^{+}: ^{40}_{19}\text{K} \to ^{40}_{18}\text{Ar} + e^{+} + \nu$
Electron capture: ^{40}_{19}\text{K} + e^{-} \to ^{40}_{18}\text{Ar} + \nu
C $\beta^{-}: ^{40}_{19}\text{K} \to ^{40}_{18}\text{Ar} + e^{-} + \bar{\nu}$
$\beta^{+}: ^{40}_{19}\text{K} \to ^{40}_{20}\text{Ca} + e^{+} + \nu$
Electron capture: ^{40}_{19}\text{K} + e^{-} \to ^{40}_{20}\text{Ca} + \nu
D $\beta^{-}: ^{40}_{19}\text{K} \to ^{40}_{20}\text{Ca} + e^{-} + \nu$
$\beta^{+}: ^{40}_{19}\text{K} \to ^{40}_{18}\text{Ar} + e^{+} + \bar{\nu}$
Electron capture: ^{40}_{19}\text{K} + e^{-} \to ^{40}_{18}\text{Ar} + \bar{\nu}

Correct answer

B. $\beta^{-}: ^{40}_{19}\text{K} \to ^{40}_{20}\text{Ca} + e^{-} + \bar{\nu}$
$\beta^{+}: ^{40}_{19}\text{K} \to ^{40}_{18}\text{Ar} + e^{+} + \nu$
Electron capture: ^{40}_{19}\text{K} + e^{-} \to ^{40}_{18}\text{Ar} + \nu

Step-by-step solution

\beta^{-}

decay, a neutron is converted into a proton, emitting an electron (

e^{-}

) and an antineutrino (

\bar{\nu}

). The atomic number increases by

1

.

^{40}_{19}\text{K} \to ^{40}_{20}\text{Ca} + e^{-} + \bar{\nu}

<br><br>In

\beta^{+}

decay, a proton is converted into a neutron, emitting a positron (

e^{+}

) and a neutrino (

\nu

). The atomic number decreases by

.<br><br>^{40}_{19}\text{K} \to ^{40}_{18}\text{Ar} + e^{+} + \nu

In electron capture, an inner orbital electron is captured by the nucleus, converting a proton into a neutron and emitting a neutrino (

\nu

). The atomic number decreases by

1

.

^{40}_{19}\text{K} + e^{-} \to ^{40}_{18}\text{Ar} + \nu

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