Atoms And Nuclei Question 408
Question: The element curium $ _{96}Cm^{248} $ has a mean life of $ 10^{13}s. $ . Its primary decay modes are spontaneous fission and $ \alpha $ -decay, the former with a probability of $ 8\% $ and the later with a probability of $ 92\% $ . Each fission releases $ 200MeV $ of energy. The masses involved in a decay are as follows: $ _{96}Cm^{248}=248.072220u, $ $ _{94}Pu^{244}=244.064100 u $ and $ _{2}He^{4}=4.002603u $ Calculate the power output from a sample of $ 10^{20}Cm $ atoms. $ (1 u=931MeV/c^{2}). $
Options:
A) $ 1.6\times {{10}^{-5}}W $
B) $ 2.6\times {{10}^{-3}}W $
C) $ 3.3\times {{10}^{-5}}W $
D) $ 5.1\times {{10}^{-3}}W $
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Answer:
Correct Answer: C
Solution:
- The total energy released E= Energy released in fission process + energy released in $ \alpha $ - decay process
$ =N_{F}\times 200+{N_{\alpha }}\times (0.005517\times 931) $ $ =( \frac{8}{100}\times 10^{20} )\times 200+( \frac{92}{100}\times 10^{20} ) $
$ \times (0.005517\times 931) $
$ =20.725\times 10^{20}MeV $ Power output $ P=E/t $
$ =\frac{20.725\times 10^{20}\times 1.6\times {{10}^{-13}}}{10^{13}} $ $ =3.3\times {{10}^{-5}}W. $
BETA
