Answered: Cadmium-101 is unstable and decays…

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1.
life of 1.4 minutes. Starting with 1.0 gram of undecayed cadmium-101, how much decay energy (in MeV)
will have been released after 2.0 minutes? The mass of a neutral atom of 101Cd is 100.918682 u, the
mass of a neutral atom of 101Ag is 100.912801 u, the mass of an electron or positron is .000549 u, and
neutrinos have no mass.
Cadmium-101 is unstable and decays according to the reaction 101Cd→ 101Ag+ e+ + ve with a half-

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Step 1

Given

Cadmium-101 is unstable and decays according to the reaction 101Cd → 101 Ag + et + ve with a half- life of 1.4 minutes. Starting with 1.0 gram of undecayed cadmium-101, how much decay energy in MeV) will have been released after 2.0 minutes. The mass of a neutral atom of 101Cd is 100.918682 u, the mass of a neutral atom of 101 Ag is 100.912801 u, the mass of an electron or positron is .000549 u, and neutrinos have no mass, as following below.

Step 2

$Here, radioactive of decay law is as follows : N = N_{0} e^{λ t} N_{0} \to initial number of nuclei / atoms Now : λ = \frac{0.693}{t_{\frac{1}{2}}} (t_{\frac{1}{2}} = 1.4 \min) \Rightarrow λ = \frac{0.693}{1.4} \Rightarrow λ = 0.495 / \min Initially there was 1 gm of {}^{101}{ca}, \Rightarrow N_{0} = \frac{6.022 \times \times 10^{23}}{101} \Rightarrow N_{0} = 5.96 \times 10^{21^{}} atoms Thus, N = 5.96 \times 10^{21^{}} \times \times e^{- 0.495 \times 2} (For t = 2 \min) N = 2.21 \times 10^{21^{}} atom$

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