13.Nuclei
hard

રેડિયો એકિટવ ન્યુકલાઈડનો ક્ષય નિયતાંક $1.5 \times 10^{-5}\,s ^{-1}$ છે. પદાર્થનો પરમાણુભાર $60\,g\,mole ^{-1},\left(N_A=6 \times 10^{23}\right)$ છે. તો $1.0 \;\mu g$ પદાર્થની એકિટવીટી $....\,\times 10^{10}\,Bq$ છે.

A

$14$

B

$13$

C

$12$

D

$15$

(JEE MAIN-2023)

Solution

$\lambda=1.5 \times 10^{-5} s ^{-1}$

No. of mole $=\frac{1 \times 10^{-6}}{60}=\frac{10^{-7}}{6}$

No. of atoms $=$ no. of moles $\times N _{ A }$ $=\frac{10^{-7}}{6} \times 6 \times 10^{23}=10^{16}$

$A=N_0 \lambda e^{-\lambda t}$

For, $t =0, A = A _0= N _0 \lambda$

$=1.5 \times 10^{-5} \times 10^{16}=15 \times 10^{10} Bq .$

Standard 12
Physics

Similar Questions

Start a Free Trial Now

Confusing about what to choose? Our team will schedule a demo shortly.