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 $

Show Answer

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. $



NCERT Chapter Video Solution

Dual Pane