Question

The mass and energy equivalent of 1amu are respectively
A. \[1.67 \times {10^{ - 27}}gm,9.30MeV\]
B. \[1.67 \times {10^{ - 27}}Kg,930MeV\]
C. \[1.67 \times {10^{ - 27}}Kg,1MeV\]
D. \[1.67 \times {10^{ - 34}}Kg,1MeV\]

Answer 1

Hint: In this we first find the mass of 1 atom using $1mole{\text{ of }}{}^{12}C = 12g = 6.023 \times {10^{23}}atoms$ and we get \[1{\text{atom}} = \dfrac{{12}}{{6.023 \times {{10}^{23}}}}g = 1.9923 \times {10^{ - 23}}g\] after that we substitute this in the equation $1amu = \dfrac{{{\text{mass of 1 atom of }}{}^{12}C}}{{12}}$ and get 1 amu equal to \[1.66 \times {10^{ - 27}}kg\]. Then using \[1amu = 1.66 \times {10^{ - 27}}kg\] and $E = m{c^2}$ we calculated $E = 1.491 \times {10^{ - 10}}J$. After that to convert Joule into MeV we used $1J = \dfrac{1}{{1.602 \times {{10}^{ - 13}}}}MeV$ and finally get $ \Rightarrow E = 931.5MeV$

Complete Step-by-Step solution:
First we need to convert 1 amu (atomic mass unit) to Kg. We already know that
$1amu = \dfrac{{{\text{mass of 1 atom of }}{}^{12}C({\text{isotope of carbon that have mass number 12)}}}}{{12}}$----------- (1)
We also know that
$1mole{\text{ of }}{}^{12}C = 12g = 6.023 \times {10^{23}}{\text{atoms of }}{}^{12}C$
Since, $6.023 \times {10^{23}}{\text{atoms}} = 12g$
We need to find the mass of 1 atom which will be the required mass of one ${}^{12}C$ atom that is
$ \Rightarrow 1{\text{atom}} = \dfrac{{12}}{{6.023 \times {{10}^{23}}}}g = 1.9923 \times {10^{ - 23}}g$---------------------------------- (2)
Therefore to get 1amu substitute equation (2) in equation (1), we get
$1amu = \dfrac{{1.9923 \times {{10}^{ - 23}}g}}{{12}}$
\[ \Rightarrow 1amu = 1.66 \times {10^{ - 24}}g = 1.66 \times {10^{ - 27}}kg\]
Hence 1 amu is \[1.66 \times {10^{ - 27}}kg\]

Now we need to convert 1 amu to mega electron volts (MeV)
Here we use the value that we have found above that is
\[1amu = 1.66 \times {10^{ - 27}}kg\]-------------------------------------------------- (3)
And also we know Einstein's famous equation relates that mass and energy that is
$E = m{c^2}$------------------------------------------------ (4)
Now substitute equation (3) in equation (4) and taking speed of light $c = 3 \times {10^8}m/s$
$E = (1.66 \times {10^{ - 27}}kg) \times {\left( {3 \times {{10}^8}m/s} \right)^2}$
$ \Rightarrow E = 1.491 \times {10^{ - 10}}J$
Here we are getting in joules but we were asked in mega electron volts (MeV). For that we know $1J = \dfrac{1}{{1.602 \times {{10}^{ - 13}}}}MeV$. Now using this we can write
$ \Rightarrow E = 1.491 \times {10^{ - 10}} \times \dfrac{1}{{1.602 \times {{10}^{ - 13}}}}MeV$
$ \Rightarrow E = 931.5MeV$
Therefore we get mass and energy equivalent of 1amu as \[1.66 \times {10^{ - 27}}kg\] and $931.5MeV$ respectively. So option B is correct.

Note: For these types of questions we need to be well versed with the concepts of moles, atomic mass number, atomic number, atomic mass unit (amu), and the standardization using the ${}^{12}C$ isotopes of carbon. We also need to have a good concept of unit conversions like g to Kg, J to MeV, etc.