$_{11}Na^{24}$ is radioactive and it decays to

  • A
    $_{9}F^{20}$ and $\alpha$-particles
  • B
    $_{13}Al^{24}$ and positron
  • C
    $_{11}Na^{23}$ and neutron
  • D
    $_{12}Mg^{24}$ and $\beta$-particles

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Similar Questions

In the radioactive disintegration series ${}_{90}^{232}Th \rightarrow {}_{82}^{208}Pb$ involving $\alpha$ and $\beta$ decay,the total number of $\alpha$ and $\beta$ particles emitted are:

The atom bomb is based on the principle of:

Consider the following nuclear reactions:
$_{92}^{238}M \to _{y}^{x}N + 2_{2}^{4}He$
$_{y}^{x}N \to _{B}^{A}L + 2\beta^{+}$
The number of neutrons in the element $L$ is:

What is $X$ in the following nuclear reaction?
$_{7}N^{14} + _{1}H^{1} \xrightarrow{} _{8}O^{15} + X$

Calculate the mass defect in the following nuclear reaction:
$_1H^2 + _1H^3 \to _2He^4 + _0n^1$
(Given: mass of $_1H^2 = 2.014 \ amu$,$_1H^3 = 3.016 \ amu$,$_2He^4 = 4.004 \ amu$,$_0n^1 = 1.008 \ amu$) (in $amu$)

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