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Chapter 18: Problem 2533

A radioactive substance decays to \((1 / 16)\) th of its initial activity in 40 days. the half life of the radioactive substance expressed in day is (A) 20 (B) 5 (C) 10 (D) 4

Short Answer

Expert verified
Given that a radioactive substance decays to \(\dfrac{1}{16}\) of its initial activity in 40 days, we first set up the radioactive decay equation: \[\dfrac{1}{16} A_{0} = A_{0}e^{-λ(40)}\] Solving for the decay constant (λ): \[λ = \dfrac{\ln \dfrac{1}{16}}{-40}\] Finally, calculating the half-life using the decay constant: \[T_{\frac{1}{2}} = \dfrac{\ln 2}{\dfrac{\ln \dfrac{1}{16}}{-40}} \] Simplifying gives us the half-life to be \(T_{\frac{1}{2}} = 10\) days. Thus, the correct answer is (C) 10.

Step by step solution

01

Given that the initial activity decays to \((1/16)\) after 40 days, the radioactive decay equation can be written as: \[ \dfrac{1}{16} A_{0} = A_{0}e^{-λ(40)}\] #Step 2: Solve the equation for the decay constant (λ)#

Divide both sides of the equation by \(A_{0}\) to get: \[ \dfrac{1}{16} = e^{-λ(40)}\] Take the natural logarithm of both sides of the equation: \[ \ln \dfrac{1}{16} = -λ(40)\] Now, we solve for \(λ\): \[ λ = \dfrac{\ln \dfrac{1}{16}}{-40} \] #Step 3: Find the half-life using the decay constant (λ)#
02

The half-life is given by the formula: \[ T_{\frac{1}{2}} = \dfrac{\ln 2}{λ} \] Use the decay constant found in Step 2 to calculate the half-life: \[ T_{\frac{1}{2}} = \dfrac{\ln 2}{\dfrac{\ln \dfrac{1}{16}}{-40}} \] #Step 4: Simplify and find the answer#

Simplify to get the half-life, and compare your answer with the given options: \[ T_{\frac{1}{2}} = 10\] So, the half-life of the radioactive substance expressed in days is \(10\). Therefore, the correct answer is (C) 10.

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Most popular questions from this chapter

The energy released by the fission of one uranium atom is \(200 \mathrm{MeV}\). The number of fission Per second required to Produce \(3.2 \mathrm{w}\) of Power is (A) \(10^{10}\) (B) \(10^{7}\) (C) \(10^{12}\) (D) \(10^{11}\)

The nucleus at rest disintegrate into two nuclear parts which have their velocities in the ratio \(2: 1\) The ratio of their nuclear sizes will be (A) \(2^{(1 / 3)}: 1\) (B) \(1: 2^{(1 / 3)}\) (C) \(3^{(1 / 2)}: 1\) (D) \(1: 3^{(1 / 2)}\)

which of the following series in the spectrum of hydrogen atom lies in the visible legion of the electro magnetic spectrum? (A) Paschen (B) Lyman (C) Brakett (D) Balmer

Match column I and II and chose correct Answer form the given below. (a) Nuclear fusion (p) converts some matter into energy (b) Nuclear fission (q) generally Possible for nuclei with low atomic number (c) \(\beta\) decay (r) generally Possible for nuclei with high atomic number (d) Exothermic nuclear (s) Essentially Proceeds by weak reaction nuclear force(c) (A) $\mathrm{a} \rightarrow \mathrm{p}, \mathrm{b} \rightarrow \mathrm{r}, \mathrm{c} \rightarrow \mathrm{s}, \mathrm{d} \rightarrow \mathrm{q}$ (B) $\mathrm{a} \rightarrow \mathrm{q}, \mathrm{b} \rightarrow \mathrm{r}, \mathrm{c} \rightarrow \mathrm{p}, \mathrm{d} \rightarrow \mathrm{s}$ (C) $\mathrm{a} \rightarrow \mathrm{q}, \mathrm{b} \rightarrow \mathrm{r}, \mathrm{c} \rightarrow \mathrm{s}, \mathrm{d} \rightarrow \mathrm{p}$ (D) $\mathrm{a} \rightarrow \mathrm{r}, \mathrm{b} \rightarrow \mathrm{q}, \mathrm{c} \rightarrow \mathrm{p}, \mathrm{d} \rightarrow \mathrm{s}$

The size of the atom is of the order of (A) \(10^{-14} \mathrm{~m}\) (B) \(10^{-10} \mathrm{~m}\) (C) \(10^{-8} \mathrm{~m}\) (D) \(10^{-6} \mathrm{~m}\)
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