Consider an electron in a hydrogen atom,revol | vedclass

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(B) For an electron in a hydrogen atom,the condition for a stable orbit according to Bohr's postulate is given by the relation $2 \pi r_n = n \lambda_

Vedclass · Accepted Answer

$9.7$

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(B) For an electron in a hydrogen atom,the condition for a stable orbit according to Bohr's postulate is given by the relation $2 \pi r_n = n \lambda_n$,where $n$ is the principal quantum number,$r_n$ is the radius of the $n^{th}$ orbit,and $\lambda_n$ is the de-Broglie wavelength.The ground state corresponds to $n=1$,the first excited state to $n=2$,and the second excited state to $n=3$.Given that the electron is in the second excited state,we have $n=3$ and the radius $r_3 = 4.65 \, \mathring{A}$.Substituting these values into the formula:$3 \lambda_3 = 2 \pi r_3$$\lambda_3 = \frac{2 \pi (4.65 \, \mathring{A})}{3}$$\lambda_3 = 2 	imes 3.14159 	imes 1.55 \, \mathring{A}$$\lambda_3 \approx 9.738 \, \mathring{A}$.Rounding to one decimal place,we get $\lambda_3 = 9.7 \, \mathring{A}$.

Consider an electron in a hydrogen atom,revolving in its second excited state (having radius $4.65 \, \mathring{A}$). The de-Broglie wavelength of this electron is .... $\mathring{A}$.

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