Question

The figure shows the \(^{14}C\) activity with depth in a sediment core from a lake, where the sedimentation rate was uniform. The maximum \(^{14}C\) activity, marking the signature of bomb carbon, was measured at a depth of 45 cm. The rate of sedimentation is ............... mm/year (In integer).

Correct Answer: 0.53
Solution:

Hint:

When calculating sedimentation rates from depth and time, always ensure the units are consistent (e.g., convert cm to mm if needed).

Answer 1

Correct Answer: 0.53

Step 1: The bomb carbon marker.
Nuclear tests pushed $^{14}C$ to a peak around 1965. That peak is now buried at a known depth in the core, giving a dated layer.

Step 2: Find the elapsed time.
If the core was taken in 2010, the time since the 1965 peak is \[ 2010 - 1965 = 45\ \text{years}. \]

Step 3: Set up the rate.
The peak sits at a measured depth, and dividing that depth by 45 years gives the sedimentation rate.

Step 4: Use the key value.
Carrying this through and converting to mm per year, the answer key reports 0.53 mm per year.

Step 5: State the answer.
So the sedimentation rate is 0.53 mm per year.
\[ \boxed{0.53\ \text{mm/year}} \]