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A donor atom in a semiconductor has a loosely bound electron. The orbit of this electron is considerably affected by the semiconductor material but behaves in many ways like an electron orbiting a hydrogen nucleus. Given that the electrons has an effective mass of $0.07 \mathrm{~m}$ (where $\mathrm{m}_{\mathrm{e}}$ is mass of the free electron) and the space in which it moves has a permittivity $13 \varepsilon_{0}$, then the radius of the electrons lowermost energy orbit will be close to (The Bohr radius of the hydrogen atom is $0.53 Å$ )
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The correct answer is:
$\mathrm100 Å$
$$
\frac{M V^{2}}{r}=\frac{K Q^{2}}{r^{2}}
$$
$\&$
$$
M V r=\frac{n h}{2 \pi}
$$
Using data given in question $\mathrm{r}=100 Å$
\frac{M V^{2}}{r}=\frac{K Q^{2}}{r^{2}}
$$
$\&$
$$
M V r=\frac{n h}{2 \pi}
$$
Using data given in question $\mathrm{r}=100 Å$
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