
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
579k+ 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.
Recently Updated Pages
Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Trending doubts
Differentiate between an exothermic and an endothermic class 11 chemistry CBSE

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Explain zero factorial class 11 maths CBSE

Why is steel more elastic than rubber class 11 physics CBSE

State the laws of reflection of light

What is 1s 2s 2p 3s 3p class 11 chemistry CBSE

