Question

A radioactive nucleus decays according to following series
$X\stackrel{\alpha }{\to }{X}_1\stackrel{{\beta }^{-}}{\to }{X}_2\stackrel{\alpha }{\to }{X}_3\stackrel{\gamma }{\to }{X}_4$
If the atomic number and atomic weight of the parent element $X$ are 72 and 180 respectively, then the atomic number and atomic mass of $X_4$ are respectively

• A. 70, 172
• B. 69, 171
• C. 69, 172
• D. 68, 172

Answer 1

The correct option is C 69, 172

We know that An alpha particle is basically a Helium nucleus. A beta (minus) particle is a high speed electron. Their mass is ~ $\frac{1}{2000}amu$ . A gamma ray is NOT a particle.

So by mass and charge balance we can write that,
${}_{72}^{180}X{\stackrel{\alpha }{\to }}_{70}^{176}{X}_1{\stackrel{{\beta }^{-}}{\to }}_{71}^{176}{X}_2{\stackrel{\alpha }{\to }}_{69}^{172}{X}_3{\stackrel{\gamma }{\to }}_{69}^{172}{X}_4$
$\therefore \left(c\right)$ is the correct option.