Obtain the amount of ${}_{27}^{60}{Co}$ necessary to provide a radioactive source of 8.0 mCi strength. The half-life of ${}_{27}^{60}{Co}$ is 5.3 years.

Solution

Given,
Half life , T = 5.3 years

Strength of radioactive source,
= 8.0 mCi = 8.0 x 10^-3 Ci
= 8.0 x 10^-3 x 3.7 x 10^-10disintegrations s^-1= 29.6 x 10⁷ disintegrations s^-1Since the strength of the source decreases with time,
$∴$ $\frac{dN}{dt} = - 29.6 \times 10^{7}$

But $\frac{dN}{dt} = - λN$

$∴$    $- λN = - 29.6 \times 10^{7}$

$\Rightarrow$    $λN = 29.6 \times 10^{7}$

$\Rightarrow$    $N = \frac{29.6 \times 10^{7}}{λ}$
$\Rightarrow$    $N = \frac{29.6 \times 10^{7} \times T}{0.693}$    $(∵ λ = \frac{0.693}{T})$
$= \frac{29.6 \times 10^{7} \times 5.3 \times 365 \times 24 \times 60 \times 60}{0.693}$

   $= 7.139 \times 10^{16}$

Number of atoms in 60 g of cobalt = 6.023 x 10²³Therefore,

Mass of 1 atom of cobalt = $\frac{60}{6.023 \times 10^{23}} g$

Mass of $7.139 \times 10^{16} atoms$ is,

$= \frac{60}{6.023 \times 10^{23}} \times 7.139 \times 10^{16} g$

$= 7.11 μg .$

For Daily Free Study Material Join wiredfaculty

Wave Optics

Obtain the amount of ${}_{27}^{60}{Co}$ necessary to provide a radioactive source of 8.0 mCi strength. The half-life of ${}_{27}^{60}{Co}$ is 5.3 years.

Some More Questions From Wave Optics Chapter

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

Phone Number

Email Address

Loaction

For Daily Free Study Material Join wiredfaculty

Wave Optics

Obtain the amount of Co2760 necessary to provide a radioactive source of 8.0 mCi strength. The half-life of Co2760 is 5.3 years.

Some More Questions From Wave Optics Chapter

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

Obtain the amount of ${}_{27}^{60}{Co}$ necessary to provide a radioactive source of 8.0 mCi strength. The half-life of ${}_{27}^{60}{Co}$ is 5.3 years.