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What fraction of the radioactive species...

What fraction of the radioactive species `(t_(1//2) = 12 "years")` remains after six years ?

A

`0.36`

B

`0.60`

C

`0.20`

D

`0.45`

Text Solution

Verified by Experts

The correct Answer is:
A
`(N_(0))/(N)= e^(lambdat), therefore N = N_(0)e^(-lambdat)" Now", alpha = (N_(0) - N)/(N_(0)) = (N_(0) - N_(0)^(-lambdat))/(N_(0)) = 1-e^(-lambdat)`
`therefore alpha = 1 - e^((0.693)/(12)) = 0.29`
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Knowledge Check

  • The activity of a radioactive samples reduces by 10% in 12.5 yr. The half-life of this radioactivity species when it is reduded to 90% .

    A
    28.20 yr
    B
    82.20 yr
    C
    2.5 yr
    D
    12.5 yr

  • The time required for a radioactive species to decay (2)/(3) of its initial amount is 't'. The fraction of radioactive species left after 0.5 t is :

    A
    `(1)/(sqrt(3))`
    B
    `(1)/(sqrt(5))`
    C
    `sqrt((2)/(3))`
    D
    `(1)/(3)`

  • The decay constant of a radioactive species is lambda for the process in which a parent element showing formation of a daughter element.After time t, P atoms of parent element ar left and D atoms of daughter elements are formed. If t_(1//2) is half life then which expression correctly represents decay of parent element,

    A
    `t = (t_(1//2))/(0.693)` In `(1 + (D)/(P))`
    B
    `t = (t_(1//2))/(0.693)` In `(1 -(D)/(P))`
    C
    `t = (t_(1//2))/(0.693)` In `((D)/(P))`
    D
    `t = (t_(1//2))/(0.693)` In `((P)/(D))`
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