Q170 / 180 · KCET 2014 · Physics
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 =
\( 1.6 \times 10^{-19} \mathrm{C} \).
\( 8.8 \times 10^{10} \mathrm{C} \mathrm{kg}^{-1} \). What is the mass of the electron ? Given charge of the electron =
\( 1.6 \times 10^{-19} \mathrm{C} \).
- A\( 1 \times 10^{-29} \mathrm{~kg} \)
- B\( 0.1 \times 10^{-29} k g \)
- C\( 1.1 \times 10^{-29} \mathrm{~kg} \)
- Correct answerD\( \frac{1}{11} \times 10^{-29} \mathrm{~kg} \)
Solution
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 \( =\frac{1}{11} \times 10^{-29} \mathrm{~kg} \)
Hence, the correct answer is option: 'D'.