An electron in an excited state of Li^{2+} io | vedclass

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(B) According to Bohr's postulate,the angular momentum $L$ is given by $L = \frac{nh}{2\pi}$.Given $L = \frac{3h}{2\pi}$,we find $n = 3$.The radius of

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$2$

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(B) According to Bohr's postulate,the angular momentum $L$ is given by $L = \frac{nh}{2\pi}$.Given $L = \frac{3h}{2\pi}$,we find $n = 3$.The radius of the $n^{th}$ orbit for a hydrogen-like ion is $r_n = a_0 \frac{n^2}{Z}$.For $Li^{2+}$,the atomic number $Z = 3$. Thus,$r_3 = a_0 \frac{3^2}{3} = 3a_0$.The de Broglie wavelength $\lambda$ is given by $\lambda = \frac{h}{mv}$.From the angular momentum formula,$mvr = \frac{nh}{2\pi}$,so $mv = \frac{nh}{2\pi r}$.Substituting this into the wavelength formula: $\lambda = \frac{h}{nh / (2\pi r)} = \frac{2\pi r}{n}$.For $n = 3$ and $r = 3a_0$,we get $\lambda = \frac{2\pi (3a_0)}{3} = 2\pi a_0$.Comparing this with $p\pi a_0$,we get $p = 2$.

An electron in an excited state of $Li^{2+}$ ion has angular momentum $3h/2\pi$. The de Broglie wavelength of the electron in this state is $p\pi a_0$ (where $a_0$ is the Bohr radius). The value of $p$ is

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