放射性物质的质量 $m$ 毫克随着时间 $t$ 小时而减少。测得的 $m$ 和 $t$ 的值见下表。

t (小时)	2	4	6	8	10
m (毫克)	48.2	41.5	35.7	30.7	26.5

已知 $m$ 和 $t$ 由方程 $m = m_0 e^{-kt}$ 相关联，其中 $m_0$ 和 $k$ 是常数。

1. Calculate the values of $\ln m$ for each given value of $m$, to 3 decimal places.

For $t = 2$, $\ln m = \ln 48.2 \approx 3.875$
For $t = 4$, $\ln m = \ln 41.5 \approx 3.726$
For $t = 6$, $\ln m = \ln 35.7 \approx 3.575$
For $t = 8$, $\ln m = \ln 30.7 \approx 3.424$
For $t = 10$, $\ln m = \ln 26.5 \approx 3.277$

2. Plot the points on the grid according to the given scales and draw the line of best fit.

Set up the horizontal axis ($t$) with 1 cm representing 1 unit, from 0 to at least 10.
Set up the vertical axis ($\ln m$) with 2 cm representing 0.1 unit, starting from 3.2.
Plot the points $(2, 3.875)$, $(4, 3.726)$, $(6, 3.575)$, $(8, 3.424)$, and $(10, 3.277)$.
Draw a straight line of best fit passing through these points.

1. Express the given equation in linear form

$m = m_0 e^{-kt}$
Taking natural logarithms on both sides:
$\ln m = \ln(m_0 e^{-kt})$
$\ln m = -kt + \ln m_0$
This is in the form $Y = mX + c$, where $Y = \ln m$, $X = t$, gradient $= -k$, and $Y$-intercept $= \ln m_0$.

2. Calculate the gradient of the line to find $k$

Using the points $(2, \ln 48.2)$ and $(10, \ln 26.5)$ from the table:
$\text{Gradient} = \frac{\ln 26.5 - \ln 48.2}{10 - 2}$
$-k = \frac{3.2771 - 3.8753}{8}$
$-k = -0.074775$
$k \approx 0.0748$ (to 3 s.f.)

3. Find the $Y$-intercept to calculate $m_0$

Using the point $(2, 3.8753)$ and the gradient $-0.074775$:
$\ln m_0 = Y - mX$
$\ln m_0 = 3.8753 - (-0.074775)(2)$
$\ln m_0 = 4.02485$
$m_0 = e^{4.02485} \approx 56.0$ (to 3 s.f.)

1. Determine the target mass after a 20% reduction

From part (ii), the initial mass $m_0$ is approximately $56.0$ mg (where $\ln m_0 \approx 4.025$).
A reduction of $20\%$ means the new mass is $80\%$ of $m_0$.
$m = 0.8 \times 56.0 = 44.8$ mg

2. Calculate the corresponding $\ln m$ value and read $t$ from the graph

$\ln m = \ln(44.8) \approx 3.80$
Using the straight line graph of $\ln m$ against $t$, locate $3.80$ on the vertical axis and read the corresponding value on the horizontal axis.
$t \approx 3.0$ hours