The wavelength of the electron in the ground | vedclass

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(A) The de Broglie wavelength is given by $\lambda = \frac{h}{mv}$.For an electron in the $n^{th}$ orbit of a hydrogen-like species,the velocity $v$ i

Vedclass · Accepted Answer

$2 y$

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(A) The de Broglie wavelength is given by $\lambda = \frac{h}{mv}$.For an electron in the $n^{th}$ orbit of a hydrogen-like species,the velocity $v$ is proportional to $\frac{Z}{n}$.Since the circumference of the orbit is $2 \pi r = n \lambda$ and $r \propto \frac{n^2}{Z}$,we have $\lambda = \frac{2 \pi r}{n} \propto \frac{n^2/Z}{n} = \frac{n}{Z}$.For the $H$-atom $(Z=1)$ in the ground state $(n=1)$: $\lambda_1 \propto \frac{1}{1} = 1$. Given $\lambda_1 = y$.For the $He^{+}$ ion $(Z=2)$ in the fourth orbit $(n=4)$: $\lambda_2 \propto \frac{4}{2} = 2$.Therefore,$\frac{\lambda_2}{\lambda_1} = \frac{2}{1} = 2$,which implies $\lambda_2 = 2 y$.

The wavelength of the electron in the ground state of a hydrogen atom is $y \ \mathring{A}$. What is the wavelength of the electron in the fourth orbit of $He^{+}$ ion (in $\mathring{A}$)?

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