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(D) The radius of the $n$-th Bohr orbit for a hydrogen-like system is given by $r_n = \frac{\varepsilon_0 n^2 h^2}{\pi m Z e^2}$.In a semiconductor,th

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$100$

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(D) The radius of the $n$-th Bohr orbit for a hydrogen-like system is given by $r_n = \frac{\varepsilon_0 n^2 h^2}{\pi m Z e^2}$.In a semiconductor,the electron moves in a medium with permittivity $\varepsilon = \varepsilon_r \varepsilon_0$ and has an effective mass $m^* = 0.07 \, m_e$.The radius of the orbit in this medium becomes $r = \frac{\varepsilon_r \varepsilon_0 n^2 h^2}{\pi m^* Z e^2}$.We know the Bohr radius of the hydrogen atom is $a_0 = \frac{\varepsilon_0 h^2}{\pi m_e e^2} = 0.53 \, \mathring{A}$.Substituting the given values $(n=1, Z=1, \varepsilon_r = 13, m^* = 0.07 \, m_e)$:$r = \frac{\varepsilon_r}{m^*/m_e} 	imes a_0 = \frac{13}{0.07} 	imes 0.53 \, \mathring{A}$.$r = \frac{6.89}{0.07} \, \mathring{A} = 98.42 \, \mathring{A}$.Rounding to the nearest integer,we get $r \approx 100 \, \mathring{A}$.

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