Paragraph: Polonium ( P ₀²¹⁰ ) emits ₂ ^4 particles and is converted into lead ( ₈₂ ~Pb ²⁰⁶ ) . This reaction is used for producing electric power in a space mission: P ₀ ²¹⁰ has half life of 138.6 days. (Given masses of the nuclei) . P ₀²¹⁰=209.98264 amu , Pb ²⁰⁶=205.97440,,^4=4.00260 amu ) Question: What is the initial amount of P ₀²¹⁰ ?

Step-by-step solution

₈₄ P ₀²¹⁰ ₈₂ ~Pb ²⁰⁶+ ₂ He ^4 Mass converted to energy per reaction is aligned & m = m ( ₈₄ P ₀²¹⁰ )- [ m ( ₈₂ ~Pb ²⁰⁶ )+ m ( ₂ He ^4 ) ] & =5.25 MeV =8.4 10⁻¹³ Joule aligned Total amount of energy required =1.2 10^7 Joule Input Energy =1.2 10^7 10=1.2 10^8 ~J Number of reaction required per day n = dN dt = 1 7 10²¹ per day If N is the number of polonium atoms required the - dN dt = N = n = N N = n = nT 0.693 = 200 7 10²¹ Mass of each P ₀- atom =210 amu Mass of P ₀ required after 693 days = ( 200 7 ) 10²¹ 210 amu =