Question

Consider a hydrogen atom with its electron in the \[{{n}^{th}}\] orbital. An electromagnetic radiation of wavelength 90 nm is used to ionize the atom. If the kinetic energy of the ejected electron is 10.4 eV, then the value of n is (hc = 1242 eV nm)

Answer 1

Hint: We are given a hydrogen atom whose electron is in the \[{{n}^{th}}\] orbital. It is said that an electromagnetic radiation is used to ionize the atom. We are given the wavelength of the radiation and the energy of the excited electron. We know that the energy of the electromagnetic radiation will be equal to the energy of the electron and the energy required for ejecting the electron. By finding these energies and equating them we will get the solution.

Formula used:
$E=\dfrac{hc}{\lambda }$
${{E}_{n}}=\dfrac{-13.6}{{{n}^{2}}}$

Complete step-by-step answer:
In the question it is said that an electromagnetic radiation ionizes a hydrogen atom in its ${{n}^{th}}$ orbital.
The wavelength of the electromagnetic radiation used is given as 90 nm, i.e.
$\lambda =90nm$
We are also given the kinetic energy of the ejected electron,
$KE=10.4eV$
We know that the energy of the 90 nm wavelength radiation is given by the equation,
$E=\dfrac{hc}{\lambda }$
In the question we are given that $hc=1242eVnm$ and we know that $\lambda =90nm$. Therefore we get the energy as,
$\Rightarrow E=\dfrac{1242}{90}$
$\Rightarrow E=13.8eV$
This energy of the light will be equal to the energy of the electron in the ${{n}^{th}}$ orbital and the kinetic energy, i.e.
$\dfrac{hc}{\lambda }=\left| {{E}_{n}} \right|+KE$
We know that the energy of electron in the ${{n}^{th}}$ for a hydrogen like atom is given as,
 ${{E}_{n}}=\dfrac{-13.6}{{{n}^{2}}}$
Therefore we get,
$\Rightarrow 13.8=\dfrac{13.6}{{{n}^{2}}}+10.4$
$\Rightarrow 13.8-10.4=\dfrac{13.6}{{{n}^{2}}}$
$\Rightarrow 3.4=\dfrac{13.6}{{{n}^{2}}}$
$\Rightarrow {{n}^{2}}=\dfrac{13.6}{3.4}$
$\Rightarrow {{n}^{2}}=4$
$\Rightarrow n=\sqrt{4}=2$
Therefore we get the value of n as 2.

Note: We know that for a hydrogen like atom the energy of electron in the \[{{n}^{th}}\] is given as,
${{E}_{n}}=\dfrac{-13.6}{{{n}^{2}}}$
 Here when we calculate, we take the modulus of this energy, i.e. we only take the magnitude and we neglect the sign.
This is done because the value of ‘n’ can never be negative.