What is the energy (in MeV) released in the alpha decay of 231Pa? July 26, 2021 by Đan Thu What is the energy (in MeV) released in the alpha decay of 231Pa?
Answer: 3729.87MeV Explanation: ²³¹₉₁Pa → ²²⁷₈₉Ac + ⁴₂He Molar mass of of He = 4.0 Number of moles = 1 Na (Avogadro’s number) = 6.023*10²³ atoms Note : 1 mole = molarmass = Avogadro’s number Molar mass = 4.000 * 1 = 4g = 4*10^-3 kg Mass = (number of moles * molar mass ) / Avogadro’s number. Mass = 4.0*10^-3 / 6.023*10^-23 Mass = 6.64 * 10^-27 kg from Einstein’s theory of relativity, energy E and mass m are interconvertible. ∇E = ∇mc² E = (6.64*10^-27) * (3.0*10⁸)² E = 5.98*10^-10J 1eV = 1.602*10-19J XeV = 5.98*10-19J X = (5.98*10^-10 * 1) / 1.602*10-19 X = 37328333958eV X = 3729.87MeV Reply
Answer: 3729.87MeV
Explanation:
²³¹₉₁Pa → ²²⁷₈₉Ac + ⁴₂He
Molar mass of of He = 4.0
Number of moles = 1
Na (Avogadro’s number) = 6.023*10²³ atoms
Note : 1 mole = molarmass = Avogadro’s number
Molar mass = 4.000 * 1 = 4g = 4*10^-3 kg
Mass = (number of moles * molar mass ) / Avogadro’s number.
Mass = 4.0*10^-3 / 6.023*10^-23
Mass = 6.64 * 10^-27 kg
from Einstein’s theory of relativity, energy E and mass m are interconvertible.
∇E = ∇mc²
E = (6.64*10^-27) * (3.0*10⁸)²
E = 5.98*10^-10J
1eV = 1.602*10-19J
XeV = 5.98*10-19J
X = (5.98*10^-10 * 1) / 1.602*10-19
X = 37328333958eV
X = 3729.87MeV