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A gyromagnetic ratio of the electron revolving in a circular orbit of hydrogen atom is
\( 8.8 \times 10^{10} \mathrm{C} \mathrm{kg}^{-1} \). What is the mass of the electron ? Given charge of the electron =
\( 1.6 \times 10^{-19} \mathrm{C} \).
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\( 8.8 \times 10^{10} \mathrm{C} \mathrm{kg}^{-1} \). What is the mass of the electron ? Given charge of the electron =
\( 1.6 \times 10^{-19} \mathrm{C} \).
Solution:
2298 Upvotes
Verified Answer
The correct answer is:
\( \frac{1}{11} \times 10^{-29} \mathrm{~kg} \)
Given, gyromagnetic ratio \( =8.8 \times 10^{10} \mathrm{C} \mathrm{kg}^{-1} \)
We know, gyromagnetic ratio \( =\frac{e}{2 m_{e}} \)
\( \Rightarrow m_{e}=\frac{e}{2 g}=\frac{1.6 \times 10^{-19}}{2 \times 8.8 \times 10^{10}}=\frac{2}{22} \times 10^{-29}=\frac{1}{11} \times 10^{-29} \mathrm{~kg} \)
Therefore, mass of electron \( =\frac{1}{11} \times 10^{-29} \mathrm{~kg} \)
We know, gyromagnetic ratio \( =\frac{e}{2 m_{e}} \)
\( \Rightarrow m_{e}=\frac{e}{2 g}=\frac{1.6 \times 10^{-19}}{2 \times 8.8 \times 10^{10}}=\frac{2}{22} \times 10^{-29}=\frac{1}{11} \times 10^{-29} \mathrm{~kg} \)
Therefore, mass of electron \( =\frac{1}{11} \times 10^{-29} \mathrm{~kg} \)
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