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Chapter 18: Problem 2462

A radioactive sample has \(\mathrm{n}_{0}\) active atom at \(\mathrm{t}=\mathrm{o}\), at the rate of disintegration at any time is \(\mathrm{R}\) and the number of atom is \(\mathrm{N}\), then ratio. $(\mathrm{R} / \mathrm{N})\( varies with time \)(\mathrm{t})$ as.

Short Answer

Expert verified
The ratio \(\frac{R}{N}\) is equal to the decay constant λ and remains constant with time. Therefore, the ratio does not vary with time.

Step by step solution

01

Radioactive Decay Law

The radioactive decay law states that the rate of decay of a radioactive substance is proportional to the number of active atoms present at any given time. Mathematically, this can be represented as: \(R = \lambda N\) where R is the rate of decay at time t, N is the number of active atoms at time t, and λ is the decay constant.
02

Exponential Decay Equation

We can also write an expression for N, the number of active atoms at any time t, in terms of the initial number of active atoms n₀: \(N = n_{0} e^{-\lambda t}\) where N is the number of active atoms at time t, n₀ is the initial number of active atoms, λ is the decay constant, and t is time.
03

Express R in terms of N

We can write the rate of decay R in terms of the number of active atoms N and the initial number of active atoms n₀ using the radioactive decay law: \(R = \lambda N\) Substitute the expression for N from Step 2: \(R = \lambda (n_{0} e^{-\lambda t})\)
04

Calculate R/N as a function of t

From the expression for R in Step 3, divide by N to find the ratio R/N as a function of time t: \(\frac{R}{N} = \frac{\lambda (n_{0} e^{-\lambda t})}{(n_{0} e^{-\lambda t})}\) Simplify, by canceling the common terms (n₀ and e⁽⁻ᴸᵗ⁾): \(\frac{R}{N} = \lambda\)
05

Conclusion

The ratio R/N is equal to the decay constant λ and remains constant with time. Therefore, the ratio does not vary with time.

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Most popular questions from this chapter

The masses of neutron and Proton are \(1.0087\) amu and \(1.0073\) amu respectively. It the neuron and Protons combins to form Helium nucleus of mass \(4.0015\) amu then binding energy of the helium nucleus will be (A) \(14.2 \mathrm{MeV}\) (B) \(28.4 \mathrm{MeV}\) (C) \(27.3 \mathrm{MeV}\) (D) \(20.8 \mathrm{MeV}\)

The control rod in a nuclear reactor is made of (A) uranium (B) cadmium (C) plutonium (D) graphite

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) \((3 / \mathrm{e})\)

The half life time of a radioactive elements of \(\mathrm{x}\) is the same as the mean life of another radioactive element \(\mathrm{y}\). Initially they have same number of atoms, then (A) \(\mathrm{y}\) will decay faster then \(\mathrm{x}\) (B) \(\mathrm{x}\) will decay faster then \(\mathrm{y}\) (C) \(\mathrm{x}\) and \(\mathrm{y}\) will decay at the same rate at all time (D) \(\mathrm{x}\) and \(\mathrm{y}\) will decay at the same rate initially.

The nuclear of which of following Pairs of nuclei are isotones (A) \({ }_{34} \mathrm{Se}^{74},{ }_{31} \mathrm{Ca}^{71}\) (B) \(_{42} \mathrm{Mo}^{92}, 40 \mathrm{Zr}^{92}\) (C) \({ }_{38} \mathrm{Sr}^{81}, 38 \mathrm{Sr}^{86}\) (D) \({ }_{20} \mathrm{Cd}^{40}, 16 \mathrm{~S}^{32}\)
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