## Related questions with answers

Question

A $1.00 \cdot 10^{-6} \ g$ sample of nobelium, $_{102}^{254} \mathrm{No}$, has a half-life of 55 seconds after it is formed. What is the percentage of $_{102}^{254} \mathrm{No}$ remaining at the following times?

(a) 5.0 min after it forms

(b) 1.0 h after it forms

Solution

VerifiedStep 1

1 of 4We have a formula from which we can calculate the ratio of the radioactive substance:

$\dfrac{N}{N_0}=e^{-\lambda {\cdot} t}$

where $t$ is time, $\lambda$ is decay constant, $N_0$ is initial concentration and $N$ is concentration at time $t$.

From the half-life we can get $\lambda$:

$\lambda= {\dfrac{ln(2)}{t_{1/2}}}$

$\lambda= {\dfrac{ln(2)}{55\text{s}}}$

$\lambda=0.0126\text{s}^{-1}$

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