You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard VII

Chemistry

Sub-Atomic Particles

The selling r...

Question

The selling rate of a radioactive isotope is decided by its activity. The second-hand rate (in rupees) of a one month old $^{32} P$ sample having $t_{1 / 2} = 14.3$ day if it was originally purchased for 800 rupees will be:

Open in App

Solution

$k = \frac{0.693}{t_{1 / 2}} = \frac{2.303}{t} l o g \frac{a}{a - x}$

Substitute values in the above expression.

$k = \frac{0.693}{14.3} = \frac{2.303}{30} l o g \frac{800}{a - x}$

$l o g \frac{800}{a - x} = 0.6313$

$\frac{800}{a - x} = 4.2785$
$a - x = 187$

Suggest Corrections

Similar questions

The selling rate of a radioactive isotope is decided by its activity. What will be the second-hand rate of a one month old $^{32} P (t_{\frac{1}{2}} = 14.3 days)$ source if it was originally purchased for 800 rupees ?

Q. The radioactive isotope

^{32} P

decays by first order kinetics and has a half-life of 14.3 days. How long does it take for 95.0% of a given sample of

^{32} P

to decay?

Q. 54.5 mg of

N a_{3} P O_{4}

contains

^{32} P

(15.6%) and

^{31} P

atoms. Assuming only

^{32} P

atoms radioactive, the rate of decay for the given sample of

N a_{3} P O_{4}

is :
(Half-life period for

^{32} P = 14.3 d a y

; molar mass of

N a_{3} P O_{4} = 161.2

)

Q. The radioactive isotope (

T_{\frac{1}{2}}

= 3days) sample was received after 12 days. It was found that 3g of the isotope is present in the sample. The initial weight of the isotope when packed is ___

Q. A radio nuclide of

^{32} P

with half-life 14.3 day is produced in a nuclear reactor at a constant rate,

q = 2.7 \times 10^{9}

nuclei per second. Period soon after the beginning of production of that nuclide will its activity be equal to

1.7 \times 10^{9}

dps is :

Join BYJU'S Learning Program

The selling rate of a radioactive isotope is decided by its activity. The second-hand rate (in rupees) of a one month old 32P sample having t1/2=14.3 day if it was originally purchased for 800 rupees will be:

k=0.693t1/2=2.303tlogaa−xSubstitute values in the above expression.k=0.69314.3=2.30330log800a−xlog800a−x=0.6313800a−x=4.2785a−x=187

The selling rate of a radioactive isotope is decided by its activity. The second-hand rate (in rupees) of a one month old $^{32} P$ sample having $t_{1 / 2} = 14.3$ day if it was originally purchased for 800 rupees will be:

$k = \frac{0.693}{t_{1 / 2}} = \frac{2.303}{t} l o g \frac{a}{a - x}$

Substitute values in the above expression.

$k = \frac{0.693}{14.3} = \frac{2.303}{30} l o g \frac{800}{a - x}$

$l o g \frac{800}{a - x} = 0.6313$

$\frac{800}{a - x} = 4.2785$
$a - x = 187$