a simple radioactive element with half life of two minutes initially contains​ 60×10^6atoms. calculate the time it would take

Question

a simple radioactive element with
half life of two minutes initially contains​ 60×10^6atoms. calculate the time it would take for the atom to reduced to 7.5×10^6atoms

in progress 0
Ngọc Khuê 4 years 2021-08-16T20:42:35+00:00 1 Answers 11 views 0

Answers ( )

  1. Dulcie
    0
    2021-08-16T20:44:09+00:00
    Reply

    Answer:

    t=360S

    Explanation:

    T½=2mins

    No=60×10^6

    t=?

    Nr=7.5×10^6

    Nr/No=1/2^n

    7.5×10^6/60×10^6=(1/2)^n

    0.125=2^- n

    Take log to bothsides

    Log10^0.125=log10^2^- n

    Ln0.125= -nln2

    – 2.079= – 0.693n

    N=2.079/0.693

    N=3

    N=t/T

    t=nxT

    t=3x2x60

    t=360s

Leave an answer

Browse

Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )