Question

The polonium isotope \[_{84}^{210}Po\] is unstable and emits 10 MeV alpha particles. The atomic mass of \[_{84}^{210}Po\] is 209.983 U and that of \[_2^4Po\] is 4.003 U. The atomic mass of the daughter nuclei is
A.210 U
B.208 U
C.82.0 U
D.None of these

Answer 1

Hint: Thus polonium-210 (mass number 210 and atomic number 84, i.e., a nucleus with 84 protons) decays by alpha emission to lead-206 (atomic number 82). Since, the atomic mass of \[_{84}^{210}Po\] is given to be 209.983 U and that of \[_2^4He\] is given to be 4.003 U. We can suppose the atomic mass of daughter nuclei is X and atomic number of daughter nuclei is Y. We can calculate the atomic mass of the daughter nuclei by subtracting atomic mass of \[_2^4He\] from \[_{84}^{210}Po\]

Complete step by step answer:
\[_{84}^{210}Po \to _Y^XA + _2^4He + 10Mev\]
Let the atomic mass of daughter nuclei is X and atomic number of daughter nuclei is Y.
\[X = 209.983-4.003 = 205.98\]
\[Y = 84-2 = 82\]

So, the atomic mass of the daughter nuclei is 206 U.

Note: The alternate way of solving this problem is
\[Mp = Md + \dfrac{E}{{931}} + Md\]
\[ \Rightarrow 210 = Md + \dfrac{{10}}{{931}} + 4.003\]
\[ \Rightarrow 210 = {M_d} + 0.010 + 4.003\]
\[ \Rightarrow {M_d} = 210 - 4.013\]
\[ \Rightarrow {M_d} = 205.987\]
Polonium-210 has a half-life of 138 days, and it decays to stable lead-206 by emitting an alpha particle (an alpha particle has two protons and two neutrons). With a specific activity of 166 TBq/g, it would take about a microgram to deliver a 50 Gy (5,000 rad) whole-body radiation dose. Polonium-210 is used mainly in static eliminators, which are devices designed to eliminate static electricity in machinery where it can be caused by processes such as paper rolling, manufacturing sheet plastics, and spinning synthetic fibres. The polonium-210 is generally electroplated onto a backing foil and inserted into a brush, tube, or other holder.