Calculating radioactivity after a period of time

[FollowTheFez]

Homework Statement

QUESTION
A cancer patient is typically given an intravenous dose of Metastron with an activity of
148 MBq.
Calculate the activity after 6 months (183 days).

PREVIOUS INFORMATION
Physical half-life of Metastron = 50.5 days
Biological half-life of Metastron = 1.8x10⁴ days

Homework Equations

T_1/2 = \frac{ln2}{λ}

\frac{ΔN}{Δt} = λN

N = N₀e^-λt

The Attempt at a Solution

The equation used to calculate the activity would be
N = N₀e^-λt

N₀ = 148MBq
t = 183days

So I just need to find λ. - This is where I think I am going wrong.

λ = \frac{ln2}{T}
λ = \frac{0.693}{50.5days}
λ = 1.59x10^-7

Substituting this into the first equation gets...

N = N₀e^-λt
N = 148 e^{-(1.59x10-7) x 183}
N = 147.996MBq
N = 148MBq

The correct answer is (apparently) 11.95MBq.
Seeing as all values are in MBq and days, I don't think I have to convert to Bq and seconds (actually even if I do convert I still get an incorrect answer).
Therefore I think I am stuffing up at my calculation of λ. But I don't see any other way in which I can calculate it. Am I correct in thinking this is where I am going wrong?

 

[pbuk]

1. You are pressing the wrong buttons on your calculator: $\frac{ln2}{50.5} \approx 0.013726$ which would give an answer of 12.01MBq.
2. You are ignoring the biological half life; treating this correctly gives an answer of 11.92MBq if 6 months is taken as 183 days, or 12.00MBq after 365/2 = 182.5 days.