Question

In the given nuclear reaction, how many a and b particles are emitted
$_{92}{X^{235}}{ \to _{82}}{Y^{207}}$
A. 3a particles and 2b particle
B. 4a particles and 3b particle
C. 6a particles and 4b particle
D. 7a particles and 4b particle

Answer 1

Hint:We know that every alpha particle (α) is a helium nucleus. Thus the alpha disintegration is shown as$_Z{X^A}{ \to _{Z - 2}}{Y^{A - 4}}$ . Where A is the atomic mass number and Z is the atomic number. Also beta () disintegration is shown as $_Z{X^A}{ \overset{\beta }{\rightarrow} _{Z + 1}}{Y^A}$. To know the alpha particle, subtract the atomic masses of both elements such as X and Y, and divide it with 4. Then for the beta particles, the formula of an atomic number of X is equal to the atomic number of $Y + 2\alpha - \beta $ .

Formula used:



Complete answer:
From our lesson we have learnt about the radioactivity, α-disintegration and β-disintegration. Alpha decay is nothing but it is a type of radioactive disintegration in which the nucleus of an atom emits an alpha particle and there is transformation or disintegration takes place. This type decays into a different atomic nucleus. And its mass number is reduced by four, and the number of atoms is reduced by two. Here $_2H{e^4}$ is the alpha particle. As we already know, helium contains two neutrons and two protons. After the emission nucleus of an atom, the mass number is reduced by four and atomic number is reduced by two.
Hence there is some change in $_Z{X^A}$ to the emitting nucleus $_{Z - 2}{Y^{A - 4}}$ is expressed as below,
$_Z{X^A}{ \to _{Z - 2}}{Y^{A - 4}}{ + _2}H{e^4}$
In beta decay, it is a type of disintegration in which the nucleus decays continuously by the emission of electrons and positrons. It has two types of decay : beta plus decay and beta minus decay. In the first type of beta plus decay there is some emission of positrons which is named as positive electrons.
+ Decay expressed as below,
$_Z{X^A}{ \to _{Z - 1}}{Y^A} + {e^ + } + \nu $
In - decay there is an emission of electrons from the nucleus. it is expressed as below,
$_Z{X^A}{ \to _{Z + 1}}{Y^A} + {e^ - } + {\nu ^ - }$
Now we need to find the alpha and beta particle in$_{92}{U^{235}}{ \to _{82}}P{b^{207}}$
It is also written in another way as,
$_{92}U^{235}\overset {{\alpha},{\beta}}{\longrightarrow} _{82}{Pb}^{207}+_{2}{He}^{4}+_{-1}e^0$
Here to find the alpha particle we could solve it by atomic mass of lead (Pb) is equal to atomic mass of uranium 4α
$(U - 4\alpha )$ ………………… (1)
Atomic mass of Uranium = 235
Atomic mass of lead = 207
Submit the both values in equation (1), we can get
207 = 235 ‒ 4α
   4α = 235 ‒207
    4α = 28
      α = 7
Now we have to find a beta particle by using the formula of the atomic number of Uranium is equal to the atomic number of $Pb + 2\alpha - \beta $. Uranium has an atomic number as 92 and lead has an atomic number 82 and we already found out the value of α.
By submitting the value which we know can get,
$U = pb + 2\alpha - \beta $
92 = 82 + 2 × 7‒β
  β = 92 ‒82 + 14
92‒82‒14+β = 0
92‒96 + β = 0
            ‒4 + β = 0
                     β = 4
Thus the number of alpha particles is 7 and the number of beta particles is 4.
Hence, the correct answer is option D


Note:The given reaction about radioactive disintegration is not a direct disintegration. Because it goes under a number of alpha and beta disintegration. The Positive electrons called positrons do not refer to protons, and it stands for neutrino and it is charged positively. Also it is a neutral particle without mass. There is no external energy in alpha decay. When an atomic mass is decreased that is caused by alpha disintegration not by beta disintegration. And beta decay can increase or decrease the atomic mass number.