Gyromagnetic ratio is the ratio of magnetic moment (μ1) to the orbital angular momentum (l) Its numerical value for an electron is given by

Question

Gyromagnetic ratio is the ratio of magnetic moment (μ1) to the orbital angular momentum (l) Its numerical value for an electron is given by (A) 8.8 × 10-12 C kg-1 (B) 8.8 × 1010 C kg-1 (C) 1.6 × 10-19 C kg-1 (D) 6.67 × 1011 C kg-1

Tardigrade Admin · Accepted Answer

The gyromagnetic ratio is a constant.
For any charged particle its value is given by

\frac{μ _{l}}{l} = \frac{e}{2 m}

For electron

m = m_{e} = 9.1 \times 1 0^{- 31} k g

,

e = 1.6 \times 1 0^{- 19} C

,

∴ \frac{μ _{l}}{l} = \frac{1.6 \times 1 0 ^{- 19}}{2 \times 9.1 \times 1 0 ^{- 31}}

= 8.8 \times 1 0^{10} C k g^{- 1}

Q. Gyromagnetic ratio is the ratio of magnetic moment $(μ_{1})$ to the orbital angular momentum $(l)$ Its numerical value for an electron is given by

$8.8 \times 1 0^{- 12} C k g^{- 1}$

$8.8 \times 1 0^{10} C k g^{- 1}$

$1.6 \times 1 0^{- 19} C k g^{- 1}$

$6.67 \times 1 0^{11} C k g^{- 1}$

Q. Gyromagnetic ratio is the ratio of magnetic moment (μ1​) to the orbital angular momentum (l) Its numerical value for an electron is given by

8.8×10−12Ckg−1

8.8×1010Ckg−1

1.6×10−19Ckg−1

6.67×1011Ckg−1

The gyromagnetic ratio is a constant. For any charged particle its value is given by lμl​​=2me​ For electron m=me​=9.1×10−31kg, e=1.6×10−19C, ∴lμl​​=2×9.1×10−311.6×10−19​ =8.8×1010Ckg−1

Q. Gyromagnetic ratio is the ratio of magnetic moment $(μ_{1})$ to the orbital angular momentum $(l)$ Its numerical value for an electron is given by

$8.8 \times 1 0^{- 12} C k g^{- 1}$

$8.8 \times 1 0^{10} C k g^{- 1}$

$1.6 \times 1 0^{- 19} C k g^{- 1}$

$6.67 \times 1 0^{11} C k g^{- 1}$