Question

Below is shown the energy-level diagram of hydrogen like imaginary element X. (hc = 1242 eV-nm)

• A. The ionization energy of Element X is 4 eV
• B. An atom in the ground state absorbs a photon, comes to an excited state then emits a photon with a wavelength of 1242 nm. The incident photon must have a wavelength of 414 nm.
• C. An atom in the ground state has a collision with an electron, then emits a photon with a wavelength of 1242 nm. The incident electron must have an energy of 3 eV.
• D. An atom in the ground state absorbs a photon, comes to an excited state then emits a photon with a wavelength of 1242 nm. The incident photon must have a wavelength of 1242 nm.

Answer 1

Answer: (A), (B)

The correct options are
A The ionization energy of Element X is 4 eV
B An atom in the ground state absorbs a photon, comes to an excited state then emits a photon with a wavelength of 1242 nm. The incident photon must have a wavelength of 414 nm.


(A) Ionization - energy = minimum – energy required to eject the $e^{-}=4\text{eV}$
(B) Energy of a photon
$E=\frac{1242\text{eV.nm}}{\lambda \left(nm\right)}$
= 1 eV
E = 1 eV corresponds to n = 3
$\lambda$ of incident photon
$\Delta E=3\text{eV}$
$\lambda =\frac{1242\text{eV.nm}}{3\text{eV}}$
$=414\text{nm}$
(C) Energy of an e– which make collision if its energy is less than excitation energy then it gets absorb

$KE{I}_{\text{system}}=\frac{1}{2}\mu v_{rel}^2.=\frac{1}{2}M{V}_{com}^2$
$3eV=\frac{1}{2}\mu v_{rel}^2.+$ constant
Hence change in K.E. of the system
$\frac{1}{2}\mu V_{rel}^2.<3\text{eV}$
Hence, incident electron must have an energy more than 3 eV