The energy of separation of an electron in a | vedclass

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(B) The energy of separation in an excited state is given by $E_n = 13.6 	imes \frac{Z^2}{n^2} \ eV = 3.4 \ eV$. From this,$\frac{Z^2}{n^2} = \frac{3

Vedclass · Accepted Answer

$6.66$

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(B) The energy of separation in an excited state is given by $E_n = 13.6 	imes \frac{Z^2}{n^2} \ eV = 3.4 \ eV$. From this,$\frac{Z^2}{n^2} = \frac{3.4}{13.6} = \frac{1}{4}$,so $\frac{n}{Z} = 2$. The radius of the $n^{th}$ orbit is $r_n = 0.53 	imes \frac{n^2}{Z} \ \mathring{A}$. According to the de-Broglie hypothesis,the circumference of the orbit is $2 \pi r_n = n \lambda$. Substituting $r_n$: $2 \pi 	imes 0.53 	imes \frac{n^2}{Z} = n \lambda$. Simplifying for $\lambda$: $\lambda = 2 \pi 	imes 0.53 	imes \frac{n}{Z}$. Substituting $\frac{n}{Z} = 2$: $\lambda = 2 	imes 3.14159 	imes 0.53 	imes 2 \approx 6.66 \ \mathring{A}$.

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

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