The gyromagnetic ratio and Bohr magneton are given respectively by [Given $\rightarrow \mathrm{e}=$ charge…

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].
  1. $\frac{\mathrm{e}}{2 \mathrm{~m}}, \frac{\mathrm{eh}}{4 \pi \mathrm{~m}}$
  2. $\frac{\mathrm{eh}}{4 \pi \mathrm{~m}}, \frac{\mathrm{e}}{2 \mathrm{~m}}$
  3. $\frac{2 \mathrm{~m}}{\mathrm{e}}, \frac{4 \pi \mathrm{~m}}{\text { eh }}$
  4. $\frac{4 \pi \mathrm{~m}}{\mathrm{eh}}, \frac{2 \mathrm{~m}}{\mathrm{e}}$

Solution

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 \(=\frac{\frac{n h}{2 \pi}}{n \times \frac{e h}{4 \pi m}}=\frac{e}{2 m}\)

Asked in: MHT CET 2024 (09 May Shift 1)

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