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Question: Find the mass of the electron in the atomic mass unit. A) 0.0005498 B) 0.5119 C) 0.5498 D) N...

Find the mass of the electron in the atomic mass unit.
A) 0.0005498
B) 0.5119
C) 0.5498
D) None of these

Explanation

Solution

The actual mass of any element is represented in the unified atomic mass unit and is denoted by u. The atomic mass unit or unified atomic mass unit is defined as one-twelfth of the actual mass of the carbon – 12 atom.

Formula used:
The atomic mass unit is given by, 1u=112×actual mass of C12atom1{\text{u}} = \dfrac{1}{{12}} \times {\text{actual mass of }}{{\text{C}}^{12}}{\text{atom}}

Complete step by step answer.
Step 1: List the mass of the electron in kg and define the atomic mass unit based on the mass of a carbon – 12 atom.
The mass of an electron in kg is 9.10938×10−31kg9.10938 \times {10^{ - 31}}{\text{kg}} .
The atomic mass unit is given by, 1u=112×actual mass of C12atom1{\text{u}} = \dfrac{1}{{12}} \times {\text{actual mass of }}{{\text{C}}^{12}}{\text{atom}} .
There are 6 protons and 6 neutrons in the carbon – 12 atom.
The molar mass of a carbon – 12 is 12.0 g/mol.
Step 2: Using the Avagadro’s number, find the actual mass of one carbon – 12 atoms.
In one mole the number of atoms present is given by the Avagadro’s number as 6.022×10236.022 \times {10^{23}} .
Therefore the actual mass of carbon – 12 is 12.06.022×1023=1.9927×10−23g=1.9927×10−26kg\dfrac{{12.0}}{{6.022 \times {{10}^{23}}}} = 1.9927 \times {10^{ - 23}}{\text{g}} = 1.9927 \times {10^{ - 26}}{\text{kg}}
Step 3: Find the value of one atomic mass unit and then find the atomic mass of the electron by unit conversion.
The actual mass of one carbon – 12 atom is 1.9927×10−26kg1.9927 \times {10^{ - 26}}{\text{kg}} .
Then 1u=112(1.9927×10−26)1{\text{u}} = \dfrac{1}{{12}}\left( {1.9927 \times {{10}^{ - 26}}} \right) i.e., 1u=1.6605×10−27kg1{\text{u}} = 1.6605 \times {10^{ - 27}}{\text{kg}} .
Since the mass of the electron in kg is 9.10938×10−31kg9.10938 \times {10^{ - 31}}{\text{kg}} , a simple unit conversion will provide the mass of the electron in u.
We get the mass of electron in u as 9.10938×10−311.6605×10−27=0.0005485u\dfrac{{9.10938 \times {{10}^{ - 31}}}}{{1.6605 \times {{10}^{ - 27}}}} = 0.0005485{\text{u}} .

Therefore, the correct option is D.

Note: Alternate method One atomic mass unit (u) can also be expressed implicitly as 1u=1NA1{\text{u}} = \dfrac{1}{{{N_A}}} where NA=6.022×1023{N_A} = 6.022 \times {10^{23}} is the Avogadro number. It describes the number of atoms present on one mole of a substance.
Then we get, 1u=16.022×1023=1.6605×10−24g1{\text{u}} = \dfrac{1}{{6.022 \times {{10}^{23}}}} = 1.6605 \times {10^{ - 24}}{\text{g}} or 1u=1.6605×10−27kg1{\text{u}} = 1.6605 \times {10^{ - 27}}{\text{kg}}
Since the mass of the electron in kg is 9.10938×10−31kg9.10938 \times {10^{ - 31}}{\text{kg}} , a simple unit conversion will provide the mass of the electron in u.
We get, mass of electron in u as 9.10938×10−311.6605×10−27=0.0005485u\dfrac{{9.10938 \times {{10}^{ - 31}}}}{{1.6605 \times {{10}^{ - 27}}}} = 0.0005485{\text{u}}
Therefore, the correct option is D.