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Chapter 18: Problem 2494

starting with a sample of Pure \({ }^{66} \mathrm{cu},(7 / 8)\) of it decays into \(\mathrm{Zn}\), 15 minutes. The half life the sample is (A) \(5 \mathrm{~min}\) (B) \(7.5 \mathrm{~min}\) (C) \(10 \mathrm{~min}\) (D) \(15 \mathrm{~min}\)

Short Answer

Expert verified
The half-life of the given sample of Cu-66 is 5 minutes, as determined by using the decay formula and solving for the half-life. The correct option is (A) \(5 \mathrm{~min}\).

Step by step solution

01

Identify the known values

We are given the following values: - Initial amount of Cu-66: \(N_0\) - Amount of Cu-66 remaining after 15 minutes: \((1/8) N_0\), because 7/8 of it decays, so \((1 - (7/8)) N_0 = (1/8) N_0\). - Time elapsed (t): 15 minutes
02

Plug the known values into the decay formula

Now we will plug these values into the decay formula: \[(1/8) N_0 = N_0 (1/2)^{15/T}\]
03

Solve for the half-life (T)

First, we will cancel \(N_0\) from both sides of the equation, since we are only interested in the half-life. \[(1/8) = (1/2)^{15/T}\] To solve for \(T\), we can take the logarithm of both sides, using the base \(1/2\). So we have: \[\log_{1/2} (1/8) = \frac{15}{T}\] The logarithm on the left side can be simplified: \[\log_{1/2} (1/2)^3 = \frac{15}{T}\] Which gives us: \[3 = \frac{15}{T}\]
04

Determine the half-life

Now we can solve for \(T\) by multiplying both sides by \(T\) and dividing both sides by 3: \[T = \frac{15}{3}\] \[T = 5\] So, the half-life of the given sample is 5 minutes. The correct option is: (A) \(5 \mathrm{~min}\)

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Most popular questions from this chapter

The binding energy Per nucleon of deuteron $\left({ }^{2}{ }_{1} \mathrm{H}\right)\( and Lielium nucleus \){ }_{2}{ }^{4}{ }_{2} \mathrm{He}$ ) is \(1.1 \mathrm{MeV}\) and \(7.0 \mathrm{MeV}\). respectively. If two deuteron react to form a single helium nucleus, the energy released is (A) \(23.6 \mathrm{MeV}\) (B) \(26.9 \mathrm{MeV}\) (C) \(13.9 \mathrm{MeV}\) (D) \(19.2 \mathrm{MeV}\)

\(\mathrm{A}\) and \(\mathrm{B}\) are two radioactive substance whose half lives are 1 and 2 years respectively. Initially \(10 \mathrm{~g}\) of \(\mathrm{A}\) and \(1 \mathrm{~g}\) of \(\mathrm{B}\) is taken. The time after which they will have same quantity remaining is (A) \(3.6\) years (B) 7 years (C) \(6.6\) years (D) 5 years

An \(\alpha\) -particle of energy \(5 \mathrm{MeV}\) is scattered though \(180^{\circ}\) by a fixed uranium nucleus. The distance of the closet approach is of the order of (A) \(10^{-8} \mathrm{~cm}\) (B) \(10^{-12} \mathrm{~cm}\) (C) \(10^{-10} \mathrm{~cm}\) (D) \(10^{-15} \mathrm{~cm}\)

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) \(10^{11}\)

A radio-active nucleus \({ }^{\mathrm{A}} \mathrm{Z} \mathrm{X}\) emits $3 \alpha$ -particles and 2 Positrons. the ratio of number of neutron to that of Protons in the final nucleus will be (A) \([(\mathrm{A}-\mathrm{Z}-8) /(\mathrm{Z}-8)]\) (B) \([(\mathrm{A}-\mathrm{Z}-12) /(\mathrm{Z}-4)]\) (C) \([(\mathrm{A}-\mathrm{Z}-4) /(\mathrm{Z}-8)]\) (D) \([(A-Z-4) /(Z-2)]\)
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