The energy of separation of an electron in a | vedclass

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(B) The energy of separation is given by $E_{	ext{sep}} = \frac{13.6 \ Z^2}{n^2} \ eV$. For a Hydrogen atom,$Z = 1$. Given $E_{	ext{sep}} = 3.4 \ eV

Vedclass · Accepted Answer

$6.66$

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(B) The energy of separation is given by $E_{	ext{sep}} = \frac{13.6 \ Z^2}{n^2} \ eV$. For a Hydrogen atom,$Z = 1$. Given $E_{	ext{sep}} = 3.4 \ eV$,we have $3.4 = \frac{13.6}{n^2}$,which gives $n^2 = 4$,so $n = 2$. According to Bohr's postulate,the circumference of the orbit is an integral multiple of the de-Broglie wavelength: $2 \pi r_n = n \lambda$. The radius of the $n^{th}$ orbit is $r_n = r_1 	imes n^2$,where $r_1 = 0.53 \ \mathring{A}$. Substituting the values: $2 \pi 	imes (0.53 	imes n^2) = n \lambda$. Simplifying for $\lambda$: $\lambda = 2 \pi 	imes 0.53 	imes n$. Since $n = 2$,$\lambda = 2 	imes 3.14159 	imes 0.53 	imes 2 \approx 6.66 \ \mathring{A}$.

The energy of separation of an electron in a $H$ atom in an excited state is $3.4 \ eV$. The de-Broglie wavelength (in $\mathring{A}$) associated with the above electron is,if the radius of the first orbit of the $H$ atom is $0.53 \ \mathring{A}$.

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