1.0 g of radioactive sodium decays to 0.25 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The radioactive decay follows the formula $N_t = N_0 	imes (1/2)^n$,where $n = t/T_{1/2}$.First,find the half-life $(T_{1/2})$: $0.25 \ g = 1.0 \

Vedclass · Accepted Answer

$32$

Vedclass · Answer

(B) The radioactive decay follows the formula $N_t = N_0 	imes (1/2)^n$,where $n = t/T_{1/2}$.First,find the half-life $(T_{1/2})$: $0.25 \ g = 1.0 \ g 	imes (1/2)^n$,which gives $(1/2)^2 = (1/2)^n$,so $n = 2$.Since $n = t/T_{1/2}$,$2 = 16 \ hours / T_{1/2}$,thus $T_{1/2} = 8 \ hours$.Now,for $48 \ g$ to become $3.0 \ g$: $3.0 = 48 	imes (1/2)^n$.$(1/2)^n = 3.0 / 48 = 1/16 = (1/2)^4$,so $n = 4$.Total time $t = n 	imes T_{1/2} = 4 	imes 8 \ hours = 32 \ hours$.

$1.0 \ g$ of radioactive sodium decays to $0.25 \ g$ in $16 \ hours$. Calculate the time required for $48 \ g$ of the same radioactive sodium to decay to $3.0 \ g$. (in $hours$)

Similar Questions

The decay of $_{92}U^{235}$ is a reaction of .......... order.

Radioactive lead $_{82}Pb^{201}$ has a half-life of $8 \ hours$. Starting from $1 \ mg$ of this isotope,how much will remain after $24 \ hours$?

$8 \, g$ 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 ....... $days$.

The half-life of radium is $1580 \ yrs$. Its average life will be:

The half-life of a radioactive isotope is $3 \ hours$. If the initial mass of the isotope is $256 \ g$,how much mass in $g$ will remain after $18 \ hours$ (in $.0$)?

Vedclass Test Series

Exam Paper Generator

Online Exam Module