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Chapter 18: Problem 2551

The Probability of survival of a radioactive nucleus for one mean life time is (A) \(1-(1 / \mathrm{e}) \mathrm{S}\) (B) \((1 / \mathrm{e})\) (C) \((2 / \mathrm{e})\) (D) \((3 / \mathrm{e})\)

Short Answer

Expert verified
The probability of survival of a radioactive nucleus for one mean lifetime is given by the radioactive decay formula, P(t) = \(e^{-\lambda t}\). By substituting the mean lifetime τ for the time t and the decay constant λ as \(\frac{1}{\tau}\), we get P(τ) = \(e^{-1}\), which simplifies to P(τ) = \(\frac{1}{e}\). Therefore, the correct answer is (B) \(\frac{1}{e}\).

Step by step solution

01

Understand Radioactive Decay Probability Formula

The probability of survival of a radioactive nucleus for a given time 't' can be given by the radioactive decay formula: P(t) = \(e^{-\lambda t}\) where P(t) is the probability of survival, λ (lambda) is the decay constant, and t is the time. The decay constant λ is related to the mean lifetime (τ) as follows: λ = \(\frac{1}{\tau}\) To find the probability of survival for one mean life time, we will substitute t with τ (mean life time) in the radioactive decay probability formula.
02

Plug-in Mean Lifetime in the Probability Formula

Now, let's plug-in τ for the time t in the probability formula: P(τ) = \(e^{-\lambda \tau}\) Since λ = \(\frac{1}{\tau}\), we can replace λ in the formula: P(τ) = \(e^{-\frac{1}{\tau} \tau}\) As τ cancels out, we are left with: P(τ) = \(e^{-1}\)
03

Simplify the Expression

Now, we can simplify the expression: P(τ) = \(\frac{1}{e}\) The probability of survival of a radioactive nucleus for one mean lifetime is \(\frac{1}{e}\). Answer: (B) \(\frac{1}{e}\)

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Most popular questions from this chapter

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) \(10^{-6} \mathrm{~m}\)

As the electron in Bohr is orbit of hydrogen atom Passes from state \(\mathrm{n}=2\) to \(\mathrm{n}=1\), the \(\mathrm{K} . \mathrm{E}\). and Potential energy changes as (A) Two fold, also two fold (B) four fold, two fold (C) four fold, also four fold (D) two fold, four fold

If the binding energy Per nucleon in \({ }_{3}^{7} \mathrm{Li}\) and ${ }^{4}{ }_{2}\( He nuclear is \)5.6 \mathrm{MeV}\( and \)7.06 \mathrm{MeV}$ respectively, then in the reaction $\mathrm{P}+{ }_{3} \mathrm{Li} \rightarrow 2\left({ }_{2}^{4} \mathrm{He}\right)$ (P here retrent Proton) energy of Proton must be (A) \(1.46 \mathrm{MeV}\) (B) \(39.2 \mathrm{MeV}\) (C) \(17.28 \mathrm{MeV}\) (D) \(28.24 \mathrm{MeV}\)

In the following nuclear fusion reaction ${ }_{1}^{2} \mathrm{H}+{ }_{1}^{3} \mathrm{H} \rightarrow{ }_{2}^{4} \mathrm{He}+{ }_{0} \mathrm{n}^{1}$ the repulsive potential energy between the two fusing nuclei is $7.7 \times 10^{-14} \mathrm{~J}$. The Temperature to which the gas must be heated is nearly (Boltzman constant \(\mathrm{K}=1.38 \times 10^{-23} \mathrm{JK}^{-1}\) ) (A) \(10^{3} \mathrm{~K}\) (B) \(10^{5} \mathrm{~K}\) (C) \(10^{7} \mathrm{~K}\) (D) \(10^{9} \mathrm{~K}\)

Match column I and II column I \(\frac{\text { column I }}{\text { fnucleus }}\) (a) size of (b) number of Proton in a nucleus column II (p) Z (q) \(10^{-15} \mathrm{~m}\) (r) \((\mathrm{A}-\mathrm{Z})\) (s) \(10^{-10} \mathrm{~m}\) 0 (c) size of Atom (d) Number of neutrons in a nucleus (A) $\mathrm{a} \rightarrow \mathrm{r}, \mathrm{b} \rightarrow \mathrm{s}, \mathrm{c} \rightarrow \mathrm{q}, \mathrm{d} \rightarrow \mathrm{p}$ (B) $\mathrm{a} \rightarrow \mathrm{q}, \mathrm{b} \rightarrow \mathrm{p}, \mathrm{c} \rightarrow \mathrm{s}, \mathrm{d} \rightarrow \mathrm{r}$ (C) $\mathrm{a} \rightarrow \mathrm{s}, \mathrm{b} \rightarrow \mathrm{r}, \mathrm{c} \rightarrow \mathrm{q}, \mathrm{d} \rightarrow \mathrm{p}$ (D) $\mathrm{b} \rightarrow \mathrm{s}, \mathrm{c} \rightarrow \mathrm{p}, \mathrm{c} \rightarrow \mathrm{q}, \mathrm{d} \rightarrow \mathrm{r}$
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