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Hint:Emission of an alpha particle causes a loss of mass of $4amu$ and a polarity loss of $ + 2$. The emission of a beta particle causes no loss of mass but a polarity loss of $ - 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 $\alpha $ emission: $\Delta M = 4(2n + 2p),\Delta Z = 2$.
Where $\Delta M$ is the loss of mass, $n$ is the mass of a neutron particle and $p$ is the mass of proton particles.
Loss of mass and a polarity loss due to beta particle $\beta $ emission: $\Delta M = 0,\Delta Z = - 1$
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 \[2n\] and \[2p\]. Applying these we get the relation $\Delta M = (2n + 2p)$ for mass change. We know that the mass of a neutron and proton is approximately $1amu$ each. Therefore, the loss in mass is equivalent to $\Delta M = (2n + 2p) = (2 + 2) = 4amu$.
The change in mass due to beta emission is $\Delta M = 0$ as they have negligible mass.
The atomic mass of the given particle is $238amu$.
Therefore, total change in mass due to alpha emission is determined by substituting these values:
$A = 238 - 4 = 234$
For atomic charge calculation:
When an alpha particle is emitted, the body from which it is emitted from losses charge equivalent to $ + 2$. Applying these we get the relation $\Delta Z = 2$ for charge loss.
The change in charge due to beta emission is $\Delta Z = - 1$.
The atomic mass of the given particle is $92$.
Therefore, total change in mass due to alpha emission is determined by substituting these values:
$Z = 92 - ( + 2 - 1) = 92 - 2 + 1 = 91$
Therefore, the final atomic mass $A$ and atomic charge $Z$ of the particle $Y$ are $234$ and $91$ 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.
Formulae used:Loss of mass and a polarity loss due to alpha particle $\alpha $ emission: $\Delta M = 4(2n + 2p),\Delta Z = 2$.
Where $\Delta M$ is the loss of mass, $n$ is the mass of a neutron particle and $p$ is the mass of proton particles.
Loss of mass and a polarity loss due to beta particle $\beta $ emission: $\Delta M = 0,\Delta Z = - 1$
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 \[2n\] and \[2p\]. Applying these we get the relation $\Delta M = (2n + 2p)$ for mass change. We know that the mass of a neutron and proton is approximately $1amu$ each. Therefore, the loss in mass is equivalent to $\Delta M = (2n + 2p) = (2 + 2) = 4amu$.
The change in mass due to beta emission is $\Delta M = 0$ as they have negligible mass.
The atomic mass of the given particle is $238amu$.
Therefore, total change in mass due to alpha emission is determined by substituting these values:
$A = 238 - 4 = 234$
For atomic charge calculation:
When an alpha particle is emitted, the body from which it is emitted from losses charge equivalent to $ + 2$. Applying these we get the relation $\Delta Z = 2$ for charge loss.
The change in charge due to beta emission is $\Delta Z = - 1$.
The atomic mass of the given particle is $92$.
Therefore, total change in mass due to alpha emission is determined by substituting these values:
$Z = 92 - ( + 2 - 1) = 92 - 2 + 1 = 91$
Therefore, the final atomic mass $A$ and atomic charge $Z$ of the particle $Y$ are $234$ and $91$ 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.
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