Ncert Exercises & Solutions

13.14 The nucleus \({ }_{10}^{23} \mathrm{Ne}\) decays by β– emission. Write down the β-decay equation and determine the maximum kinetic energy of the electrons emitted. Given that:

m (\({ }_{10}^{23} \mathrm{Ne}\)) = 22.994466 u

m (\({ }_{11}^{23} \mathrm{Ne}\)) = 22.089770 u.

Hint: The maximum kinetic energy of the electrons is equal to the Q-value.
Step 1: Write down the β-decay equation.
In

β^{-}

emission, the number of protons increases by 1, and one electron and an antineutrino are emitted from the parent nucleus.

β^{-}

emission of the nucleus

{}_{10}^{23}N e

{}_{10}^{23}N e \to {}_{11}^{23}N a + e^{-} + \bar{v} + Q

Step 2: Find the Q-value of the reaction.
It is given that:

The atomic mass of

m ({}_{10}^{23}{Ne})

=22.994466 u

The atomic mass of

m ({}_{11}^{23}{Na})

=22.99770 u

Mass of an electron,

m_{e}

=0.00058 u

Q-value of the given reaction is given as:

Q= [m ({}_{11}^{23}{Ne}) - [m ({}_{11}^{23}{Na}) + m_{e}]] c^{2}

There are 10 electrons in \({ }_{10}^{23} \mathrm{Na}\) and 11 electrons in

{}_{11}^{23}N a

. Hence, the mass of the electron is cancelled in the Q-value equation.

therfore, Q=[22.994466-22.989770]

c^{2}

=(0.004696

c^{2}

) u

But 1 u=931.5 Mev/

c^{2}

∴ ∆ Q = 0.004696 \times 931.5 = 4.374 M e V

Step 3: Find the maximu kinetic energy of the electrons emitted.
The daughter nucleus is too heavy as compared to

e^{-} a n d \bar{v}

. Hene, it carries negligible energy. The kinetic energy of the antineutrino is nearly zero. Hence, the maximum kinetic energy of the emitted electrons is almost equal to the Q-value, i.e., 4.374 Me.

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