The ratio of magnetic dipole moment of an electron of charge ‘e’ and mass ‘m’ in Bohr’s orbit in hydrogen atom to its angular momentum is

Question

The ratio of magnetic dipole moment of an electron of charge ‘e’ and mass ‘m’ in Bohr’s orbit in hydrogen atom to its angular momentum is (A) (e/m) (B) (m/e) (C) (2m/e) (D) (e/2m)

Tardigrade Admin · Accepted Answer

Angular momentum, L=me V r ...(i) Therefore, orbital motion of electron is equivalent to a current, I=e(1 / T) Period of revolution of electron is given by T=(2 π r/v) ∴ I=e((1/2 π r / . v))=(e v/2 π r) Area of electron orbit, A=π r2 Magnetic dipole moment of the atom M=(e V/2 π r) × π r2=e v r Using Eq. (i), we have M=((e/2 me)) L

Q. The ratio of magnetic dipole moment of an electron of charge ‘ $e$ ’ and mass ‘ $m$ ’ in Bohr’s orbit in hydrogen atom to its angular momentum is

$\frac{e}{m}$

$\frac{m}{e}$

$\frac{2 m}{e}$

$\frac{e}{2 m}$

Q. The ratio of magnetic dipole moment of an electron of charge ‘e’ and mass ‘m’ in Bohr’s orbit in hydrogen atom to its angular momentum is

me​

em​

e2m​

2me​

Q. The ratio of magnetic dipole moment of an electron of charge ‘ $e$ ’ and mass ‘ $m$ ’ in Bohr’s orbit in hydrogen atom to its angular momentum is

$\frac{e}{m}$

$\frac{m}{e}$

$\frac{2 m}{e}$

$\frac{e}{2 m}$