MHT CET · Physics · Magnetic Properties of Matter
The gyromagnetic ratio and Bohr magneton are given respectively by [Given \(\rightarrow \mathrm{e}=\) charge on electron, \(\mathrm{m}=\) mass of electron, \(\mathrm{h}=\) Planck's constant].
- A \(\frac{\mathrm{e}}{2 \mathrm{~m}}, \frac{\mathrm{eh}}{4 \pi \mathrm{~m}}\)
- B \(\frac{\mathrm{eh}}{4 \pi \mathrm{~m}}, \frac{\mathrm{e}}{2 \mathrm{~m}}\)
- C \(\frac{2 \mathrm{~m}}{\mathrm{e}}, \frac{4 \pi \mathrm{~m}}{\text { eh }}\)
- D \(\frac{4 \pi \mathrm{~m}}{\mathrm{eh}}, \frac{2 \mathrm{~m}}{\mathrm{e}}\)
Answer & Solution
Correct Answer
(A) \(\frac{\mathrm{e}}{2 \mathrm{~m}}, \frac{\mathrm{eh}}{4 \pi \mathrm{~m}}\)
Step-by-step Solution
Detailed explanation
As we know the formula gyromagnetic ratio \(=\frac{L}{M}\)
\(\text {angular momentum }(\mathrm{L})=\frac{n h}{2 \pi}\)
Magnetic moment(M) \(=n \times\) Bohrmagneton \(=n \times \frac{e h}{4 \pi m}\)
So, gyromagnetic ratio \(=\frac{\frac{n h}{2 \pi}}{n \times \frac{e h}{4 \pi m}}=\frac{e}{2 m}\)
\(\text {angular momentum }(\mathrm{L})=\frac{n h}{2 \pi}\)
Magnetic moment(M) \(=n \times\) Bohrmagneton \(=n \times \frac{e h}{4 \pi m}\)
So, gyromagnetic ratio \(=\frac{\frac{n h}{2 \pi}}{n \times \frac{e h}{4 \pi m}}=\frac{e}{2 m}\)
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