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Question: In the disintegration series \({^{238}_{92}}U{ \to ^\alpha }X{ \to ^{{\beta ^{ - 1}}}}{^A_Z}Y\) Th...

In the disintegration series 92238UαXβ1ZAY{^{238}_{92}}U{ \to ^\alpha }X{ \to ^{{\beta ^{ - 1}}}}{^A_Z}Y
The values of ZZ and AA are

A) 92,32692,326
B) 88,23088,230
C) 90,23490,234
D) 91,23491,234

Explanation

Solution

Emission of an alpha particle causes a loss of mass of 4amu4amu and a polarity loss of +2 + 2. The emission of a beta particle causes no loss of mass but a polarity loss of 1 - 1. By balancing the mass and charge we will find the atomic weight and mass of the final particle.

Formulae used: Loss of mass and a polarity loss due to alpha particle α\alpha emission: ΔM=4(2n+2p),ΔZ=2\Delta M = 4(2n + 2p),\Delta Z = 2.
Where ΔM\Delta M is the loss of mass, nn is the mass of a neutron particle and pp is the mass of proton particles.

Loss of mass and a polarity loss due to beta particle β\beta emission: ΔM=0,ΔZ=1\Delta M = 0,\Delta Z = - 1

Step by step solution:
For atomic mass calculation:
When an alpha particle is emitted, the body from which it is emitted loses a mass equivalent to 2n2n and 2p2p. Applying these we get the relation ΔM=(2n+2p)\Delta M = (2n + 2p) for mass change. We know that the mass of a neutron and proton is approximately 1amu1amu each. Therefore, the loss in mass is equivalent to ΔM=(2n+2p)=(2+2)=4amu\Delta M = (2n + 2p) = (2 + 2) = 4amu.
The change in mass due to beta emission is ΔM=0\Delta M = 0 as they have negligible mass.
The atomic mass of the given particle is 238amu238amu.
Therefore, total change in mass due to alpha emission is determined by substituting these values:
A=2384=234A = 238 - 4 = 234
For atomic charge calculation:
When an alpha particle is emitted, the body from which it is emitted from losses charge equivalent to +2 + 2. Applying these we get the relation ΔZ=2\Delta Z = 2 for charge loss.
The change in charge due to beta emission is ΔZ=1\Delta Z = - 1.
The atomic mass of the given particle is 9292.
Therefore, total change in mass due to alpha emission is determined by substituting these values:
Z=92(+21)=922+1=91Z = 92 - ( + 2 - 1) = 92 - 2 + 1 = 91
Therefore, the final atomic mass AA and atomic charge ZZ of the particle YY are 234234 and 9191 respectively.

In conclusion, the correct option is D.

Note: An alpha particle is a helium nuclei that has been stripped of its electrons. Therefore, it only consists of two proton and neutron particles. A beta particle is nothing but a high speed electron. Therefore it has negligible mass and a unit negative charge, just like an electron.