Q#1 Please show full working solution. Thesolution presented earlier for this pr

Question

Q#1   Please show full working solution. Thesolution presented earlier for this problem wasn't clear sothat is why i re-sent it for more clearification. please helpme.Thank you Initially there are 6*10 power 9 radioactive ofelement w (decay constant 0.173/day or per day) in asample. As a nucleus of w decays, it converts 2.67 *10power-28 kg of its mass to energy and releases the energy. How much time(in days) is required for the sample to release a total of7.2*10power -4 j? (a) 0 (b) 0.000290 (c) 0.00571 (d) 0.00943 (e)0.0137 (a,b) 0.0290 (a.c) 0.0571 (a,d) 0.290 (a,e) 0.571 (b,c) 3.29 Q#1   Please show full working solution. Thesolution presented earlier for this problem wasn't clear sothat is why i re-sent it for more clearification. please helpme.Thank you Initially there are 6*10 power 9 radioactive ofelement w (decay constant 0.173/day or per day) in asample. As a nucleus of w decays, it converts 2.67 *10power-28 kg of its mass to energy and releases the energy. How much time(in days) is required for the sample to release a total of7.2*10power -4 j? (a) 0 (b) 0.000290 (c) 0.00571 (d) 0.00943 (e)0.0137 (a,b) 0.0290 (a.c) 0.0571 (a,d) 0.290 (a,e) 0.571 (b,c) 3.29

Explanation / Answer

N = 6*109, = 0.173/day, m = 2.67*10-28kg, E = 7.2*10-4 J, find t initial number of nucleus = N after t, number = Ne-t number reduced = N(1 - e-t) for each nucleus, energy created = mc2 after t, energy generated = mc2N(1 -e-t) = E t = (-1/)ln[1 - E/(mc2N)] = 0.029 days answer: (a,b)

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