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Starting with a sample of pure .^(66)Cu,...

Starting with a sample of pure `.^(66)Cu, 7//8` of it decays into `Zn` in `15` minute. The corresponding half-life is:

A

`10` minute

B

`15` minute

C

`5` minute

D

`7(1)/(2)`minute

Text Solution

Verified by Experts

The correct Answer is:
C
`N=N_(0)(1-e^(-lambdat))`
`rArr (N_(0)-N)/(N_(0))=e^(-lambdat)`
`:. (1)/(8)=e^(-lambdat)`
`rArr 8=e^(lambdat t)" "rArr 3 ln 2=lambda t " "rArr=(3xx0.693)/(15)`
`t_(1//2)=(0.693)/(3xx0.693)xx15" "t_(1//2)=5 min`
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Knowledge Check

  • Starting with a sample of pure .^(66)Cu, 1/8 of it decays into Zn in 15 minutes. The corresponding half life is

    A
    `(7)1/2` minutes
    B
    5 minutes
    C
    15 minutes
    D
    10 minutes

  • Starting with a sample of pure .^66 Cu, 7//8 of it decays into Zn in 15 min . The corresponding half-life is.

    A
    10 min
    B
    5 min
    C
    15 min
    D
    `7 (1)/(2) min`

  • Starting with a sample of pure ^66Cu , 3/4 of it decays into Zn in 15 minutes. The corresponding half-life is

    A
    (a) 5 minutes
    B
    (b) 7.5 minutes
    C
    (c) 10 minutes
    D
    (d) 3.5 minutes

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