In the following nuclear reaction,
$$D\buildrel \alpha \over \longrightarrow {D_1}\buildrel {{\beta ^ - }} \over \longrightarrow {D_2}\buildrel \alpha \over \longrightarrow {D_3}\buildrel \gamma \over \longrightarrow {D_4}$$
Mass number of D is 182 and atomic number is 74. Mass number and atomic number of D4 respectively will be _________.
Solution
<p>Equivalent reaction can be written as</p>
<p>$D\buildrel {} \over
\longrightarrow {D_4} + 2\alpha + {\beta ^ - } + \gamma$</p>
<p>$\Rightarrow$ Mass number of D<sub>4</sub> = Mass number of D $-$ 2 $\times$ 4</p>
<p>= 182 $-$ 8 = 174</p>
<p>$\Rightarrow$ Atomic number of D<sub>4</sub></p>
<p>= Atomic number of D $-$ 2 $\times$ 2 + 1</p>
<p>= 74 $-$ 4 + 1 = 71</p>
About this question
Subject: Physics · Chapter: Atoms and Nuclei · Topic: Radioactivity
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