[Timeless Thinker]

Radiochemists studied copper-61 and found that 7.85×10^–5mol emitted 1.47 × 10¹⁹ positrons in 90.0 minutes. What is the half-life (in hours) for copper-61?

7.85x10^-5-(1.47×10¹⁹/6.02x10²³) = 7.85x10^-5 x (1/2)^t/HL

5.41x10^-5=7.85x10^-5 x (1/2)^90/HL

.689=(1/2)^90/HL

log(.689)=90/HL x log(1/2)

.537 = 90/HL

HL= 167.6 minutes

Is this the correct approach?

[Borek]

Logic looks OK at first sight.

[Babcock_Hall]

I don't understand why you used 1/2 as the base.  Why don't you write out the rate law for radioactive decay?

[Borek]

[itex]\left(\frac 1 2\right)^{\frac t {T_{\frac 1 2}}}[/itex] is just a direct application of the half time definition, isn't it?

[mjc123]

To be pedantic, the question asks for the half life in hours.
Be careful about things like this, it could lose you unnecessary marks in an exam.

[Babcock_Hall]

I worked the problem differently and obtained a half-life of 167.4 minutes = 2.790 hours.  This is slightly less than what I have found in quick, on-line searches (3.33 hours).  The slight difference between this problem and what I found on-line might be because of the limited amount of data in the present problem.

[AWK]

Problem with this data is solved to decay constant (problem #2, finally gives the same half-time)
https://www.chemteam.info/Kinetics/WS-Kinetics-first-order-radioactive-decay.html

[Babcock_Hall]

AWK,

The solution at the link which you provided also how I solved it.  The equation given by the OP was unfamiliar to me, which threw me off-track at first.  Thanks.

[Borek]

It helps to write what OP did in symbols, not in numbers. Then it is obvious it is a direct application of the half life definition.