Question

${}_{92}^{235}U{+}_0^{1}n{⟶}_{56}^{139}Ba{+}_{36}^{94}Kr+3_o^{1}n+200MeV$. Total energy released (in MeV) after $5^{th}$ stage of fission is

• A. 48600
• B. 16200
• C. 24200
• D. None of these

Answer 1

The correct option is C 24200

$E_{total}=200\frac{(3^5-1)}{3-1}$

$=100\times 242=24200MeV$ {Option C}