Activity of substance after a certain time

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Homework Help Overview

The discussion revolves around the activity of a radioactive substance after a specified time period, involving calculations related to half-life and decay rates. The subject area includes concepts from nuclear physics and exponential decay.

Discussion Character

  • Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the calculation of activity using the formula A = A_0 (0.5)^x, questioning the accuracy of their computations and the provided answer. There is a focus on correcting a numerical error in the half-life value used in the calculations.

Discussion Status

Participants are actively engaging in verifying their calculations and discussing the implications of the initial activity versus the activity after 48 hours. Some guidance is provided regarding the correction of the half-life value, and there is an ongoing dialogue about the correctness of both the given answer and their own calculations.

Contextual Notes

There is a noted confusion regarding the half-life value used in the calculations, which affects the outcome. Participants are also grappling with the distinction between initial activity and the activity after the elapsed time.

desmond iking
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Homework Statement


A sample consists of 2g of a radioactive element. The molar mass of the element is 67g. If the half life of the element is 78 hours, the activity of the sample after 48 hours is ?

Homework Equations

The Attempt at a Solution


d(N)/dt = -λN
(2/67 x 6.02x10^23) (ln 2 )/ (78x 3600) = 4.4x10^16s^-1
A= acitivity
A = A_0 (0.5)^x
= (4.4x10^16) x (0.5)^(48/72)[/B]
= 2.7x10^16s^-1

but the ans given is 4.4x10^16s^-1
 
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desmond iking said:
= (4.4x10^16) x (0.5)^(48/72)
78, not 72.
 
After changing 72 to 78.. Is my ans correct?
 
desmond iking said:
After changing 72 to 78.. Is my ans correct?
You posted that the answer given is 4.4x10^16s^-1, but that, as you showed, is the initial activity.
With the 72 changed to 78 your working looks right to me.
 
So the ans given is correct? Or my working is correct?
 
desmond iking said:
So the ans given is correct? Or my working is correct?
The answer given is clearly wrong because it is the initial activity. It must be less after 48 hours.
 
the ans should be
A = A_0 (0.5)^x
= (4.4x10^16) x (0.5)^(48/78)
= 2.87x10^16 ?
 
desmond iking said:
the ans should be
A = A_0 (0.5)^x
= (4.4x10^16) x (0.5)^(48/78)
= 2.87x10^16 ?
Looks right.
 

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