Magnetic dipole moment corresponding to motion of electron in second excited state of $H e^{+}$ ion is

\frac{e h}{4 m π}

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\frac{3 e h}{4 m π}

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\frac{e h}{2 m π}

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\frac{5 e h}{4 m π}

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Solution

The correct option is B $\frac{3 e h}{4 m π}$
$M = \frac{q}{2 m} L = \frac{q}{2 m} {\frac{n h}{2 π}}$
$= \frac{e}{2 m} {\frac{n h}{2 π}}$ (second excited state n = 3)
$= \frac{3 e h}{4 m π}$

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Similar questions

H e^{+}

Calculate the magnetic dipole moment corresp onding to the motion of the electron in the ground state of a hydogen atom.

μ_{1}

3^{r d}

μ_{2}

{L i}^{2 +}

(n_{2})

48.24 n m

142.46 n m

n_{2}

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