Question

The atomic mass of lead is $208\mathrm{g}$ and atomic number is $82$. The atomic mass of bismuth is $209\mathrm{g}$ and atomic number is $83$. Which one having higher $n/p$ ratio?

Answer 1

Step 1:

Definition: The ratio of number of neutrons in the atomic nucleus of an atom to the number of protons present in atomic nucleus. It is denoted by $n/p$.

Atomic mass of bismuth is $209\mathrm{g}$.

Atomic number of bismuth is 83.

Calculate the $n/p$ ratio of Bismuth ($\mathrm{Bi}$):

$$\begin{gathered} n=209-83 \\ n=126;p=83 \\ \frac{\mathrm{n}}{\mathrm{p}}=\frac{126}{83} \\ \frac{n}{p}=1.51 \end{gathered}$$

Step 2: Calculate he $n/p$ ratio of Lead ($\mathrm{Pb}$)-

Atomic mass of Lead is $208\mathrm{g}$.

Atomic number of lead is 82.

$$\begin{gathered} \mathrm{n}=208-82 \\ \mathrm{n}=126;\mathrm{p}=82 \\ \frac{\mathrm{n}}{\mathrm{p}}=\frac{126}{82} \\ \frac{\mathrm{n}}{\mathrm{p}}=1.53 \end{gathered}$$

Hence, the $n/p$ ratio of lead ($1.53$) is higher than the $n/p$ ratio of Bismuth ($1.51$).