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(A) For a hydrogen-like atom,the radius of the $n^{th}$ orbit is given by $r_n \propto n^2$. Given $r_2 = x$,we have $r_n = k \cdot n^2$. So,$x = k \c

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$2 \pi x$

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(A) For a hydrogen-like atom,the radius of the $n^{th}$ orbit is given by $r_n \propto n^2$. Given $r_2 = x$,we have $r_n = k \cdot n^2$. So,$x = k \cdot (2)^2 = 4k$,which implies $k = x/4$. For the $4^{th}$ orbit,$r_4 = k \cdot (4)^2 = 16k = 16 \cdot (x/4) = 4x$. According to the Bohr quantization condition,$mvr_n = n \frac{h}{2 \pi}$. For the $4^{th}$ orbit,$mvr_4 = 4 \frac{h}{2 \pi}$. The de-Broglie wavelength is given by $\lambda = \frac{h}{mv}$. Rearranging the quantization condition,$\frac{h}{mv} = \frac{2 \pi r_4}{n}$. Substituting $n = 4$ and $r_4 = 4x$,we get $\lambda = \frac{2 \pi (4x)}{4} = 2 \pi x$.

The radius of the second Bohr orbit is $x$. The de-Broglie wavelength of an electron in the $4^{th}$ orbit is nearly

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