(B) The total time elapsed from February $1$ to July $1$ is $28 + 31 + 30 + 31 + 30 = 150 \, days$.Using the radioactive decay formula: $N = N_0 (1/2)

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(B) The total time elapsed from February $1$ to July $1$ is $28 + 31 + 30 + 31 + 30 = 150 \, days$.Using the radioactive decay formula: $N = N_0 (1/2)^n$,where $n$ is the number of half-lives.$0.25 = 8 \times (1/2)^n$$(1/2)^n = 0.25 / 8 = 1 / 32 = (1/2)^5$.Therefore,$n = 5$.Since $n = \text{total time} / t_{1/2}$,we have $5 = 150 / t_{1/2}$.$t_{1/2} = 150 / 5 = 30 \, days$.

Class 12 Chemistry Nuclear Chemistry Rate of decay and Half-life

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$8 \, g$ of the radioactive isotope,cesium$-137$ were collected on February $1$ and kept in a sealed tube. On July $1$,it was found that only $0.25 \, g$ of it remained. So the half-life period of the isotope is ....... $days$.

A
$37.5$
B
$30$
C
$25$
D
$50$

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Nuclear Chemistry

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Rate of decay and Half-life

In the case of a radioisotope,the value of $T_{1/2}$ and $\lambda$ are identical in magnitude. The value is

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$2.25$ $C^{14}$ atoms dissociate from $1 \ g$ of carbon in an old statue. $15.3$ $C^{14}$ atoms dissociate from carbon in a living statue. How many years old is the statue? (Half-life of $C^{14} = 5730 \ \text{years}$).

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The half-life of a radioactive isotope is $3 \ hours$. If the initial mass of the isotope were $256 \ g$,the mass of it remaining undecayed after $18 \ hours$ would be ....... $g$. (in $.0$)

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Half-life of a radioactive substance which disintegrates by $75\%$ in $60 \ min$ will be ......... $min$.

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If $n_t$ number of radioatoms are present at time $t$,the following expression will be a constant:

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$8 \, g$ of the radioactive isotope,cesium$-137$ were collected on February $1$ and kept in a sealed tube. On July $1$,it was found that only $0.25 \, g$ of it remained. So the half-life period of the isotope is ....... $days$.

Similar Questions

In the case of a radioisotope,the value of $T_{1/2}$ and $\lambda$ are identical in magnitude. The value is

$2.25$ $C^{14}$ atoms dissociate from $1 \ g$ of carbon in an old statue. $15.3$ $C^{14}$ atoms dissociate from carbon in a living statue. How many years old is the statue? (Half-life of $C^{14} = 5730 \ \text{years}$).

The half-life of a radioactive isotope is $3 \ hours$. If the initial mass of the isotope were $256 \ g$,the mass of it remaining undecayed after $18 \ hours$ would be ....... $g$. (in $.0$)

Half-life of a radioactive substance which disintegrates by $75\%$ in $60 \ min$ will be ......... $min$.

If $n_t$ number of radioatoms are present at time $t$,the following expression will be a constant:

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