Question

Find the ratio of magnetic moment of electron in $He^+$ ion in ground state to the magnetic moment of electron in H atom in 1st excited state.

• A. 1
• B. 2
• C. $\frac{1}{4}$
• D. $\frac{1}{2}$

Answer 1

Answer: (D)

$\frac{1}{2}$

Answer 2

The magnetic moment ($\mu$) of an electron in an orbit is given by $\mu = \frac{eL}{2m}$, where $e$ is the electron's charge, $L$ is its orbital angular momentum, and $m$ is its mass. According to Bohr's model, the orbital angular momentum is quantized as $L_n = n\hbar$, where $n$ is the principal quantum number and $\hbar$ is the reduced Planck constant. Substituting $L_n$ into the magnetic moment formula, we get $\mu_n = \frac{e(n\hbar)}{2m} = n \left(\frac{e\hbar}{2m}\right)$. The term $\frac{e\hbar}{2m}$ is the Bohr magneton ($\mu_B$), so $\mu_n = n\mu_B$. This shows that the magnetic moment is directly proportional to the principal quantum number $n$.

For the $He^+$ ion in the ground state ($n=1$): $\mu_{He^+} = 1 \times \mu_B = \mu_B$.

For the H atom in the 1st excited state ($n=2$): $\mu_H = 2 \times \mu_B = 2\mu_B$.

The ratio is: $\text{Ratio} = \frac{\mu_{He^+}}{\mu_H} = \frac{\mu_B}{2\mu_B} = \frac{1}{2}$