This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.
ScanSolve AI Answer
Step 1: Convert the mass of the electron from atomic mass units (a.m.u.) to kilograms (kg). Given mass of electron (m) = 5.486 × 10^-4 \, a.m.u. The standard conversion factor for a.m.u. to kg is 1 \, a.m.u. = 1.660539 × 10^-27 \, kg. m = (5.486 × 10^-4 \, a.m.u.) × (1.660539 × 10^-27 \, kg/a.m.u.) m = 9.10938 × 10^-31 \, kg Step 2: Calculate the energy in Joules (J) using Einstein's mass-energy equivalence formula. The formula is E = mc^2, where: E is the energy m is the mass in kilograms c is the speed of light in a vacuum (3.0 × 10^8 \, m/s) Substitute the values: E = (9.10938 × 10^-31 \, kg) × (3.0 × 10^8 \, m/s)^2 E = (9.10938 × 10^-31) × (9.0 × 10^16) \, J E = (9.10938 × 9.0) × (10^-31 × 10^16) \, J E = 81.98442 × 10^-15 \, J E = 8.198442 × 10^-14 \, J Step 3: Convert the energy from Joules to Mega-electron Volts (MeV). Given conversion factor: 1 \, MeV = 1.602 × 10^-13 \, J. To convert Joules to MeV, divide by the conversion factor: E_MeV = E_J1.602 × 10^-13 \, J/MeV E_MeV = 8.198442 × 10^-14 \, J1.602 × 10^-13 \, J/MeV E_MeV = (8.198442)/(1.602) × 10^-1410^-13 \, MeV E_MeV = 5.117629 × 10^-1 \, MeV E_MeV = 0.5117629 \, MeV Rounding to three significant figures (consistent with the given mass and speed of light): E_MeV ≈ 0.512 \, MeV The energy of the electron is 0.512 MeV.