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The decay constant of a radioactive subs...

The decay constant of a radioactive substance is 0.173 `("years")^(-1)`. Therefore:

A

Nearly `63%` of the radioactive sustance will decay in `(1//0.173)` year

B

Half life of the radio active substance is `(1//0.173)` years

C

One-fourth of the radioactive substance is `(1//0.173)` years

D

All the above statements are true

Text Solution

Verified by Experts

The correct Answer is:
A, C

`lambda=0.173, N=N_(0) e^(-lambdat)`
`DeltaN=(N_(0) -N_(0)e^(-lambdat))=` Decayed amount
`DeltaN=N_(0) (1-1/e)`
`DeltaN=N_(0)(1-0.37)=N_(0) (0.63)`
`(Delta N)/N_(0)=[(N_(0)(0.63))/N_(0)]xx100=63%`
`T_(1//2)=(log_(e)2)/(lambda)=4` year
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Knowledge Check

  • The decay constant of a radioactive substance is 0.173 (year)""^(-1) . Therefore

    A
    nearly 63% of the radioactive substance will decay in `(1/0.173years)`
    B
    half life of the radio active substance is `(1/0.173years)`
    C
    One sixth of the radioactive substance will be left after 8 year
    D
    all the above statements are true
  • The decay constant of a radioactive substance is 0.173 year^(-1) . Therefore,

    A
    nearly `63%` of the radioactive substnce will decay in `(1//0.173)` year.
    B
    half-life of the radioactive substance is `(1//0.173)` year
    C
    One-fourth of the radioactive substance will be left after 8 years
    D
    All of the above
  • The decay constant of a radioactive sample

    A
    decreases as the atom becomes older
    B
    increases as the atom becomes older
    C
    is independent of the age
    D
    dependes on the nature of activity
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