The radioactive sources A and B with half lives of t hr and 2t hr respectively, initially contain the same number of radioactive atoms. At the end of

Question

The radioactive sources A and B with half lives of t hr and 2t hr respectively, initially contain the same number of radioactive atoms. At the end of

The radioactive sources A and B with half lives of t hr and 2t hr respectively, initially contain the same number of radioactive atoms. At the end of t hours, their rates of disintegration are in the ratio :
A. `2sqrt(2) :1`
B. `1 : 8`
C. `sqrt(2) : 1`
D. ln `2 : 1`

Monjur Alahi

5 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Correct Answer - C
`{:(,A=(t_(1//2)=t" hour")/(lambda_(1))X " ",, B=(t_(t//2)=2t" hour")/(lambda_(2)),),("at " t=0,,,,),("No. of nuclei"," "N_(0),," "N_(0),),(,"Nuclei of A left after time "t(t=t_(1//2)),,"Nuclei of B left after time "t,),(,=(N_(0))/(2),,=N_(0)e^(-lambda_(2)t),),(,"Rate of disintegration of A",,=N_(0)e^(-(ln2)/(2 t)t),),(,=lambda_(1)(N_(0))/(2),,=N_(0)e^(ln((1)/(2))^(1//2)),),(,=(ln 2)/(t) (N_(0))/(2),,=(N_(0))/(sqrt(2)),),(,,,"Rate of disintegration of B",),(,,,=lambda_(2)(N_(0))/(sqrt(2))=(ln2)/(2t)(N_(0))/(sqrt(2)),):}`
Ratio of rate of disintegration of A & B `=((ln2)/(t)(N_(0))/(2))/((ln2)/(2t)(N_(0))/(sqrt(2)))=(sqrt(2) )/(1)`