Question

One mole of radium has an activity of 1/3.7 kilo curie. Its decay constant will be​

Answer 1

Answer:

The decay constant is $1.66\times 10^{-11}\text{ s}^{-1}$

Explanation:

If A is activity in becquerel (Bq),  λ is decay constant in  per sec and N is the number of undecayed nuclei then

$\boxed{A=\lambda N}$

1 curie = 3.7 × 10¹⁰ Bq

∴ 1/3.7 kilo curie = 10¹³ Bq = A

For one mole of radium, the number of atoms = 6.023 × 10²³ = N

Therefore, decay constant

$\lambda=\frac{A}{N}$

$\implies \lambda=\frac{10^{13}}{6.023\times 10^{23}}$

$\implies \lambda=1.66\times 10^{-11}\text{ s}^{-1}$

Hope this helps.