Mass of an electron is:
(A) \[\text{9}\text{.1083 }\times \text{ 1}{{\text{0}}^{-31}}kg\]
(B) \[\text{9}\text{.1083 }\times \text{ 1}{{\text{0}}^{-24}}kg\]
(C) \[\text{9}\text{.1083 }\times \text{ 1}{{\text{0}}^{-28}}kg\]
(D) \[\text{1}\text{.67 }\times \text{ 1}{{\text{0}}^{-24}}gm\]
Answer
609.6k+ views
Hint: The mass of an electron was determined on the basis of Einstein’s Mass –Energy equivalence Principle. An electron is a negatively charged sub-particle in an atom. According to J.J Thomson’s experiment , the mass of an electron is about $\dfrac{1}{1837}$ times the mass of a proton.
Complete step by step answer:
- We will see how Sir J.J Thomson found the mass of an electron.
- By Milikan’s Oil-drop experiment, the charge of an electron (e) was measured to be \[-1.60\times {{10}^{-19}}\] Coulombs.
- Then, the mass of an electron was calculated by putting the value of the charge of an electron in the \[\dfrac{e}{m}\] ratio as found out by Sir J.J Thomson.
- The mass of the electron is calculated as follows:
- Sir J.J Thomson through his experiments, found that the ratio of the charge of an electron to the mass of an electron is equal to $-1.76\times {{10}^{8}}Cg{{m}^{-1}}$
Thus, we can write that
\[\dfrac{e}{m}=-1.76\times {{10}^{8}}\]
Now, we can write it as \[m=\dfrac{e}{-1.76\times {{10}^{8}}}\] ........(1)
But we know that charge of an electron is $-1.60\times {{10}^{-19}}C$. So, we will put this value into equation (1)
\[m=\dfrac{-1.60\times {{10}^{-19}}}{-1.76\times {{10}^{8}}}\]
So, we can simplify this as
\[m=9.1083\times {{10}^{-28}}gm\]
Now, we need to find this value in kg units.
So, we know that 1000gm = 1kg (Here, k stands for kilo which means ${{10}^{3}}$ )
So, we can write that \[9.1083\times {{10}^{-28}}gm=9.1083\times {{10}^{-28}}\times {{10}^{-3}}kg\] =\[9.1083\times {{10}^{-31}}kg\]
Thus we can say that mass of an electron is \[\text{9}\text{.1083 }\times \text{ 1}{{\text{0}}^{-31}}kg\].
So, the correct answer is “Option A”.
Note: Remember that $\dfrac{e}{m}$ ratio is the charge to mass ratio of an atom. It is essential for finding the mass of an electron. Make sure that you are taking care about and fully aware of the units you use. You need to remember the charge on an electron in order to find its mass.
Complete step by step answer:
- We will see how Sir J.J Thomson found the mass of an electron.
- By Milikan’s Oil-drop experiment, the charge of an electron (e) was measured to be \[-1.60\times {{10}^{-19}}\] Coulombs.
- Then, the mass of an electron was calculated by putting the value of the charge of an electron in the \[\dfrac{e}{m}\] ratio as found out by Sir J.J Thomson.
- The mass of the electron is calculated as follows:
- Sir J.J Thomson through his experiments, found that the ratio of the charge of an electron to the mass of an electron is equal to $-1.76\times {{10}^{8}}Cg{{m}^{-1}}$
Thus, we can write that
\[\dfrac{e}{m}=-1.76\times {{10}^{8}}\]
Now, we can write it as \[m=\dfrac{e}{-1.76\times {{10}^{8}}}\] ........(1)
But we know that charge of an electron is $-1.60\times {{10}^{-19}}C$. So, we will put this value into equation (1)
\[m=\dfrac{-1.60\times {{10}^{-19}}}{-1.76\times {{10}^{8}}}\]
So, we can simplify this as
\[m=9.1083\times {{10}^{-28}}gm\]
Now, we need to find this value in kg units.
So, we know that 1000gm = 1kg (Here, k stands for kilo which means ${{10}^{3}}$ )
So, we can write that \[9.1083\times {{10}^{-28}}gm=9.1083\times {{10}^{-28}}\times {{10}^{-3}}kg\] =\[9.1083\times {{10}^{-31}}kg\]
Thus we can say that mass of an electron is \[\text{9}\text{.1083 }\times \text{ 1}{{\text{0}}^{-31}}kg\].
So, the correct answer is “Option A”.
Note: Remember that $\dfrac{e}{m}$ ratio is the charge to mass ratio of an atom. It is essential for finding the mass of an electron. Make sure that you are taking care about and fully aware of the units you use. You need to remember the charge on an electron in order to find its mass.
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