The atomic mass of lead is 208 and atomic num | vedclass

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(A) For lead $(_{82}Pb^{208})$: Number of neutrons $(n)$ = $208 - 82 = 126$. Number of protons $(p)$ = $82$. The ratio $n/p = \frac{126}{82} \approx 1

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Higher for lead

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(A) For lead $(_{82}Pb^{208})$: Number of neutrons $(n)$ = $208 - 82 = 126$. Number of protons $(p)$ = $82$. The ratio $n/p = \frac{126}{82} \approx 1.536$. For bismuth $(_{83}Bi^{209})$: Number of neutrons $(n)$ = $209 - 83 = 126$. Number of protons $(p)$ = $83$. The ratio $n/p = \frac{126}{83} \approx 1.518$. Comparing the two,$1.536 > 1.518$,therefore the $n/p$ ratio is higher for lead.

The atomic mass of lead is $208$ and atomic number is $82$. The atomic mass of bismuth is $209$ and atomic number is $83$. The ratio of $n/p$ in the atom is

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