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  • Euler Number Solution

    Solution: Carbon-14 Dating

    1) First, rearrange the decay function:

    N = N 0 e − k t

    N N 0 = e − k t

    By substituting
    N N 0 = 1 2
    and rearranging the equation, we get:

    k = ln 2 t 1 2

    Then, substitute
    t 1 2
    = 5730 ,

    k = ln 2 5730

    k = 1.2097 × 10 − 4 y − 1

    ∴ The decay constant is 1.2097 × 10−4 y−1.

    2)

    N N 0 = e − k t

    0.4 = e − 1.2097 × 10 − 4 × t

    ln 0.4 = −1.2097 × 10−4 × t

    t = 7575y

    ∴ The age of that piece of wood is 7575 y.

    (Source: Science Focus, The Hong Kong University of Science and Technology)

  • Euler Number Solution

    答案: 碳-14 定年法

    1) 首先,重新排列衰減函數:

    N = N 0 e − k t

    N N 0 = e − k t

    代入
    N N 0 = 1 2
    並重新排列方程,得到:

    k = ln 2 t 1 2

    接著代入
    t 1 2
    = 5730,

    k = ln 2 5730

    k = 1.2097 × 10 − 4 y − 1

    ∴ 衰減常數為 1.2097 × 10−4 年−1。

    2)

    N N 0 = e − k t

    0.4 = e − 1.2097 × 10 − 4 × t

    ln 0.4 = −1.2097 × 10−4 × t

    t = 7575y

    ∴ 木材的年齡為 7575 年。

    (資料來源: 香港科技大學《科言》)

  • Euler Number Solution

    答案: 碳-14 定年法

    1) 首先,重新排列衰减函数:

    N = N 0 e − k t

    N N 0 = e − k t

    代入
    N N 0 = 1 2
    并重新排列方程,得到:

    k = ln 2 t 1 2

    接着代入
    t 1 2
    = 5730 ,

    k = ln 2 5730

    k = 1.2097 × 10 − 4 y − 1

    ∴ 衰减常数为 1.2097 × 10−4 年−1。

    2)

    N N 0 = e − k t

    0.4 = e − 1.2097 × 10 − 4 × t

    ln 0.4 = −1.2097 × 10−4 × t

    t = 7575y

    ∴ 木材的年龄为 7575 年。

    (资料来源: 香港科技大学《科言》)

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