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 $L = \frac{n h}{2 \pi}$.Given $L = \frac{3 h}{2 \pi}$, we get $n = 3$.The de-B

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$2$

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(B) According to Bohr's postulate, the angular momentum $L$ is given by $L = \frac{n h}{2 \pi}$.Given $L = \frac{3 h}{2 \pi}$, we get $n = 3$.The de-Broglie wavelength $\lambda$ is given by $\lambda = \frac{h}{mv}$.From the angular momentum quantization, $mvr = \frac{nh}{2\pi} = \frac{3h}{2\pi}$, so $mv = \frac{3h}{2\pi r}$.Substituting this into the wavelength formula: $\lambda = \frac{h}{mv} = \frac{h \cdot 2\pi r}{3h} = \frac{2}{3} \pi r$.The radius of the $n$-th orbit for a hydrogen-like ion is $r = a_{0} \frac{n^2}{Z}$.For $Li^{2+}$, $Z = 3$ and $n = 3$, so $r = a_{0} \frac{3^2}{3} = 3 a_{0}$.Substituting $r$ into the expression for $\lambda$: $\lambda = \frac{2}{3} \pi (3 a_{0}) = 2 \pi a_{0}$.Comparing this with the given form $p \pi a_{0}$, we find $p = 2$.

$A$. Speed of revolution of electron	$i$. $\frac{1}{4 \pi \varepsilon_0} \frac{2 \pi Z e^2}{n h}$
$B$. Kinetic energy	$ii$. $-\left(\frac{1}{4 \pi \varepsilon_0}\right)^2 \frac{2 \pi^2 m e^4 Z^2}{n^2 h^2}$
$C$. Total energy	$iii$. $\left(\frac{1}{4 \pi \varepsilon_0}\right)^2 \frac{2 \pi^2 m e^4 Z^2}{n^2 h^2}$
$D$. Frequency	$iv$. $\left(\frac{1}{4 \pi \varepsilon_0}\right)^2 \frac{4 \pi^2 Z^2 e^4 m}{n^3 h^3}$

An electron in an excited state of $Li^{2+}$ ion has angular momentum $\frac{3 h}{2 \pi}$. The de-Broglie wavelength of the electron in this state is $p \pi a_{0}$ (where, $a_{0} = \text{Bohr radius}$). The value of $p$ is

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