
The ${}_{90}^{232}Th$ disintegrates to ${}_{82}^{208}Pb$. How many $\beta $ particles are evolved?
Answer
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Hint: We have to remember that the beta particles are electrons (electrons with negative electric charge).Beta decay occurs when in a nucleus with too many protons or too many neutrons, one of the protons or neutrons is transformed into the other.
Complete answer:
We must know that the rapidly moving electrons emitted by nuclei are called beta particles and denoted as ${_{ - 1}^0}\beta $ because $\beta $ particle has no protons or neutrons; its mass number is zero.
The decay of nucleus by emitting a beta $\left( {{e^ - }} \right)$ particle is called as beta emission. In beta emission, one neutron of the original nucleus decays to form a beta particle and a proton. Due to which, the new nucleus formed has one more proton and one fewer neutron than the original nucleus.
The general equation is given as,
${}_Z^AX \to {}_{ - 1}^0e + {}_{\left( {Z + 1} \right)}^AY$
Where,
$A$ is the mass number.
$Z$ is the atomic number.
$X$ is the original isotope.
$Y$ is the new isotope.
Based on the question, thorium disintegrates to lead along with the emission of an alpha particle and beta particle. So, the chemical reaction is written as,
${}_{90}^{232}Th \to {}_{82}^{208}Pb + x{}_2^4\alpha + y{}_{ - 1}^0\beta $
Let x be the number of alpha particles.
Let y be the number of beta particles.
So let us solve for x,
$232 = 208 + 4x + y \times 0$
$4x = 232 - 208$
$4x = 24$
On simplifying we get,
$ \Rightarrow x = 6$
So, the number of alpha particles evolved is six.
So let us solve for y,
$90 = 82 + 12 - y$
$90 = 94 - y$
$y = 94 - 90$
On simplifying we get,
$ \Rightarrow y = 4$
So, the number of beta particles evolved is four.
So, we can write the completed nuclear equation as,
${}_{90}^{232}Th \to {}_{82}^{208}Pb + 6{}_2^4\alpha + 4{}_{ - 1}^0\beta $
The number of beta particles involved is four.
Note:
We can observe that radioactivity takes place when the nucleus of an atom that is unstable loses energy by transmitting energy in a type of electromagnetic waves (or) produced particles known as radiation. In straightforward words, the immediate emission of radiation as high energy photons (or) particles ascending from atomic response is known as radioactivity. Radioactive decay, nuclear disintegration, nuclear decay are different expressions of radioactivity.
Complete answer:
We must know that the rapidly moving electrons emitted by nuclei are called beta particles and denoted as ${_{ - 1}^0}\beta $ because $\beta $ particle has no protons or neutrons; its mass number is zero.
The decay of nucleus by emitting a beta $\left( {{e^ - }} \right)$ particle is called as beta emission. In beta emission, one neutron of the original nucleus decays to form a beta particle and a proton. Due to which, the new nucleus formed has one more proton and one fewer neutron than the original nucleus.
The general equation is given as,
${}_Z^AX \to {}_{ - 1}^0e + {}_{\left( {Z + 1} \right)}^AY$
Where,
$A$ is the mass number.
$Z$ is the atomic number.
$X$ is the original isotope.
$Y$ is the new isotope.
Based on the question, thorium disintegrates to lead along with the emission of an alpha particle and beta particle. So, the chemical reaction is written as,
${}_{90}^{232}Th \to {}_{82}^{208}Pb + x{}_2^4\alpha + y{}_{ - 1}^0\beta $
Let x be the number of alpha particles.
Let y be the number of beta particles.
So let us solve for x,
$232 = 208 + 4x + y \times 0$
$4x = 232 - 208$
$4x = 24$
On simplifying we get,
$ \Rightarrow x = 6$
So, the number of alpha particles evolved is six.
So let us solve for y,
$90 = 82 + 12 - y$
$90 = 94 - y$
$y = 94 - 90$
On simplifying we get,
$ \Rightarrow y = 4$
So, the number of beta particles evolved is four.
So, we can write the completed nuclear equation as,
${}_{90}^{232}Th \to {}_{82}^{208}Pb + 6{}_2^4\alpha + 4{}_{ - 1}^0\beta $
The number of beta particles involved is four.
Note:
We can observe that radioactivity takes place when the nucleus of an atom that is unstable loses energy by transmitting energy in a type of electromagnetic waves (or) produced particles known as radiation. In straightforward words, the immediate emission of radiation as high energy photons (or) particles ascending from atomic response is known as radioactivity. Radioactive decay, nuclear disintegration, nuclear decay are different expressions of radioactivity.
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