The wavelength of the electron in the ground state of hydrogen atom is y $Å$. What is the wavelength of the…
The wavelength of the electron in the ground state of hydrogen atom is y $Å$. What is the wavelength of the electron in the fourth orbit of $\mathrm{He}^{+}$ion (in $Å$ ) is
$2 \mathrm{y}$
$3 y$
$\mathrm{y}$
$\frac{3 \mathrm{y}}{2}$
Solution
The wavelength of an electron in the nth energy level of a hydrogen-like atom (an atom with only one electron) is given by:
$
\lambda=h / p
$
where $h$ is Planck's constant and $p$ is the momentum of the electron. The momentum of the electron is related to its energy and mass by the formula:
$
p=\operatorname{sqrt}(2 \mathrm{mE})
$
where $m$ is the mass of the electron and $E$ is its energy. The energy of an electron in the nth energy level of a hydrogen-like atom is given by:
$
\mathrm{E}=-\left(m e^{\wedge} 4\right) /\left(8 \varepsilon_0 \wedge 2 h^{\wedge} 2\right) *\left(1 / n^{\wedge} 2\right)
$
where me is the mass of the electron, $e$ is the elementary charge, $\varepsilon_0$ is the permittivity of free space, and $h$ is Planck's constant.
For the hydrogen atom, the wavelength of the electron in the ground state ( $n=1$ ) is given by:
$
\lambda=h / p=h / \operatorname{sqrt}(2 m E)=h / \operatorname{sqrt}\left(2 m *\left(-\left(m e^{\wedge} 4\right) /\left(8 \varepsilon_0 \wedge 2 h^{\wedge} 2\right)\right)\right)=h /\left(m e^{\wedge} 2 \varepsilon_0\right)
$
For the fourth orbit of a hydrogen-like ion (such as $\mathrm{He}+$ ), the energy of the electron is:
$
\mathrm{E}=-\left(m e^{\wedge} 4\right) /\left(8 \varepsilon_0 \wedge 2 h^{\wedge} 2\right) *\left(1 / 4^{\wedge} 2\right)
$
The wavelength of the electron in this orbit is:
$
\lambda^{\prime}=h / p^{\prime}=h / \operatorname{sqrt}\left(2 m E^{\prime}\right)=h / \operatorname{sqrt}\left(2 m^*\left(-\left(m e^{\wedge} 4\right) /\left(8 \varepsilon_0 \wedge 2 h^{\wedge} 2\right)\right)^*\left(1 / 4^{\wedge} 2\right)\right)=h /\left(m e^{\wedge} 2 \varepsilon_0 * 2\right)
$
The ratio of the wavelength in the fourth orbit to that in the ground state is:
$
\lambda^{\prime} / \lambda=\left(h /\left(m e^{\wedge} 2 \varepsilon_0 * 2\right)\right) /\left(h /\left(m e^{\wedge} 2 \varepsilon_0\right)\right)=1 / 2
$
So the wavelength of the electron in the fourth orbit of He+ ion is half of the wavelength in the ground state of the hydrogen atom. Therefore, the answer is (d) $3 y / 2$