Question

An element x decays, first by position emission and then two alpha particles are emitted in successive radioactive decay. If the product has a mass number 229 and atomic number 89, the mass number and atomic number of element X are.

1. 237, 93
2. 237, 94
3. 221, 84
4. 237, 92

Answer 1

Apply the concept of alpha decay and positron emission. There would be two equations formed one for the positron emission and other for the alpha decay; they both are related to each other. Solve the two equations and we will get the value of element X.

Formula used: Positron emission
$_Z^A{\rm X} \to Y + {e^ + } + \upsilon$
Here,
${\rm X}$= Element of which we have to find atomic number and mass;
A = Atomic mass;
Z = Atomic number;
Y = Product of the element after positron emission;
${e^ + }$= Positron;
$\upsilon$= Neutrino;

Equation for Alpha Decay:
$_Z^AY \to {}_{89}^{229}Z + 2\alpha$;
Here,
Y = Product of the element after positron emission;
A = Atomic mass;
Z = Atomic number;
$\alpha$= Alpha particle;
$Z$= Product of the element of Y after alpha decay;

Complete step by step solution:
Write both of the equations and solve:
Positron emission:
$_Z^A{\rm X} \to Y + e + \upsilon$;
Alpha decay,
$_Z^AY \to {}_{89}^{229}Z + 2\alpha$;
Put the values in the above two equations and solve:
$_Z^AY \to {}_{89}^{229}Z + 2\alpha$;
Here, the alpha particle ($\alpha$) has mass number 4.
$Z = 229 + 2 \times 4$;
The atomic number of alpha particles ($\alpha$) is 2.
$A = 89 + 2 \times 2$;
Do the needed calculation,
$Z = 229 + 8 = 237$;
$A = 89 + 4 = 93$;
Now put the above value of atomic number and mass number in the equation given below for Y.
$_Z^A{\rm X} \to {}_{93}^{237}Y + {e^ + } + \upsilon$;
Solve and find the value of X:
According to the above equation we have to add the atomic mass and number of positrons (${e^ + }$) and neutrino ($\upsilon$).
Here the atomic mass of positron and neutrino is zero.
$A = 237 + 0 + 0 = 237$;
Similarly, the atomic number of positron is 1 and neutrino is zero.
$Z = 93 + 1 + 0 = 94$;
So, X will have the atomic mass and number:
$_Z^A{\rm X} \to {}_{94}^{237}{\rm X}$;
Final Answer: If the product has a mass number 229 and atomic number 89, the mass number and atomic number of element X are $_Z^A{\rm X} \to {}_{94}^{237}{\rm X}$; A = 237; Z = 94. Therefore option “2” is correct.

Note: Here the alpha particle’s atomic number and mass is identical to the atomic number and mass of the helium atom. Here we have taken the atomic number and mass of the alpha particle to be 2 and 4 respectively. There are two equations that are formed $_Z^A{\rm X} \to Y + e + \upsilon$ & $_Z^AY \to {}_{89}^{229}Z + 2\alpha$. Solve the latter one first and then move on to the first equation.