Question

Each deuterium atom is formed from 1 proton, 1 neutron, and 1 electron. How much energy is released by the formation of 1 mole of deuterium (D)?

Useful Info:
• E = mc² Mass defect: 0.00238 amu
• 1 amu = 1.6606 x 10-²7 kg
• c=3.00 x 10 m/s
• 1 mol D = 6.022 x 1023 D atoms
A. 2.14 x 10¹¹ J/mol
B. 3.56 x 10-⁻¹³J/mol
C. 2.14 x 10¹⁴ J/mol
D. 1,29 x 10³⁸ J/mol

Answer 1

The energy released by the formation of one mole of deuterium atoms is calculated by multiplying the nuclear binding energy per atom by Avogadro's number, giving a total of 1.29 x 10³¸ J/mol.

Step-by-step explanation:

The question asks about the energy released by the formation of one mole of deuterium atoms. Each deuterium atom consists of 1 proton, 1 neutron, and 1 electron. The nuclear binding energy for a deuterium nucleus, which indicates the stability of the nucleus, can be expressed both in terms of the mass defect and in terms of energy. For deuterium, the mass defect is 0.002388 atomic mass units (amu), which corresponds to a nuclear binding energy of 2.22 million electron volts (MeV). To calculate the energy for one mole of deuterium, we multiply the nuclear binding energy per atom by Avogadro's number (6.022 x 1023 atoms/mol).

Since 1 MeV is equivalent to 1.60218 x 10-13 joules (J), the energy released in joules per mole can be determined by converting 2.22 MeV to joules and then multiplying by Avogadro's number. Using these conversions, we find that the energy released by the formation of one mole of deuterium atoms is 1.29 x 1038 J/mol.