Question:

The relation between  \(λ\) and \(T_\frac 12\)is: 
\((T_\frac 12 = half\  life, \ λ→decay\  constant)\)

Updated On: Oct 10, 2023
  • \(T_{\frac 12}=\frac {ln\ 2}{λ}\)

  • \(T_{\frac12} ln\ 2=λ\)

  • \(T_{\frac12}=\frac 1λ\)

  • \((λ+T_{\frac 12})=\frac {ln\ 2}{2}\)

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The Correct Option is A

Solution and Explanation

The decay of atoms over time can be described using the formula \(N_t = N_0 . e^{(-λt)}\)
where
\(N_t\) represents the number of atoms at time \(t\) 
\(N_0\) is the initial number of atoms, 
\(λ\) is the decay constant.
For a specific case where \(t\) equals \(T_\frac12\):
\(2N_o = N_0 . e^{(-λ .T_\frac12)}\)
Simplifying this expression, we find:
\(2 = e^{(-λ . T_\frac  12)}\)
Taking the natural logarithm (ln) of both sides, we get:
\(ln\ (2) = -λ . T_\frac 12\)
Now, solving for \(T_\frac12\)
\(T_\frac 12 = \frac {ln\ (2)}{λ}\)
This equation establishes the relationship between the decay constant \((λ)\) and the half-life \((T_\frac 12)\).

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Concepts Used:

Nuclei

In the year 1911, Rutherford discovered the atomic nucleus along with his associates. It is already known that every atom is manufactured of positive charge and mass in the form of a nucleus that is concentrated at the center of the atom. More than 99.9% of the mass of an atom is located in the nucleus. Additionally, the size of the atom is of the order of 10-10 m and that of the nucleus is of the order of 10-15 m.

Read More: Nuclei

Following are the terms related to nucleus:

  1. Atomic Number
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  3. Nuclear Size
  4. Nuclear Density
  5. Atomic Mass Unit