
$_{92}^{235}\text{U}$ atom disintegrates $_{82}^{207}\text{Pb}$ with a half life of ${{10}^{9}}$ years. In the process it emits 7 alpha particles and n$\beta $ particles. Here n is
a) 7
b) 3
c) 4
d) 14
Answer
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Hint: The radioactive atoms of the respective elements disintegrate such that the atoms emit alpha particles, beta particles as well as electrons. When an atom emits alpha or Beta particles, there is a certain amount of change in the atomic mass as well as the atomic number. Hence knowing how much is the change caused by a single emission of alpha and beta particles, accordingly we can determine how many particles are emitted by knowing the change in the atomic mass as well as the change in atomic number after disintegration.
Complete step-by-step answer:
When a radioactive element undergoes disintegration, the respective atoms of the elements emit alpha ($\alpha $ )and the beta ($\beta $ )particle. On disintegration an emission of a single alpha particle reduces the atomic number by 2 and atomic mass by 4. On release of a single beta particle the atomic number gets increased by one.
In the question it is given that $_{92}^{235}\text{U}$ atom disintegrates into$_{82}^{207}\text{Pb}$. The change in the atomic number as a result of above disintegration is 28. Since a single alpha particle reduces the atomic mass by 4, 7 alpha particles are emitted. Since seven alpha particles are emitted and on emission of a single alpha particle the atomic number reduces by 2, the change in the atomic number as a result of above disintegration will be 14 resulting in the atomic number of Pb to be 78. But the atomic number of Pb is 82. On emission of a single beta particle the atomic number gets increased by 1. Hence the number of beta particles emitted is 4.
So, the correct answer is “Option c)”.
Note: On emission of Beta particles there is no change in the atomic mass of the respective atoms of the element. The atomic mass of the element is given by the superscript and the atomic number of the respective element is given by the subscript. It is also to be noted that electrons as well as neutrons are also emitted in atomic disintegration.
Complete step-by-step answer:
When a radioactive element undergoes disintegration, the respective atoms of the elements emit alpha ($\alpha $ )and the beta ($\beta $ )particle. On disintegration an emission of a single alpha particle reduces the atomic number by 2 and atomic mass by 4. On release of a single beta particle the atomic number gets increased by one.
In the question it is given that $_{92}^{235}\text{U}$ atom disintegrates into$_{82}^{207}\text{Pb}$. The change in the atomic number as a result of above disintegration is 28. Since a single alpha particle reduces the atomic mass by 4, 7 alpha particles are emitted. Since seven alpha particles are emitted and on emission of a single alpha particle the atomic number reduces by 2, the change in the atomic number as a result of above disintegration will be 14 resulting in the atomic number of Pb to be 78. But the atomic number of Pb is 82. On emission of a single beta particle the atomic number gets increased by 1. Hence the number of beta particles emitted is 4.
So, the correct answer is “Option c)”.
Note: On emission of Beta particles there is no change in the atomic mass of the respective atoms of the element. The atomic mass of the element is given by the superscript and the atomic number of the respective element is given by the subscript. It is also to be noted that electrons as well as neutrons are also emitted in atomic disintegration.
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