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# The energy level diagram below represents the different energy states for a particular atom. What is the energy of a photon which is associated with the shortest possible wavelength emitted by this atom?${\text{A}}{\text{. 16 eV}}$${\text{B}}{\text{. 8 eV}}$${\text{C}}{\text{. 13 eV}}$${\text{D}}{\text{. 15 eV}}$${\text{E}}{\text{. 1 eV}}$

Last updated date: 08th Sep 2024
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Hint: The energy of the photon is inversely proportional to the wavelength and thus, directly proportional to the frequency. This is because wavelength and frequency are inversely proportional to each other. As the wavelength increases, the energy of the photon decreases.

Formula used: $E = \dfrac{{hc}}{\lambda }$
Where, E is the energy of the photon,
$h$ is Planck's constant,
$c$ is the velocity of light in vacuum,
$\lambda$ is the wavelength.

$E = \dfrac{{hc}}{\lambda }$
$\Rightarrow E \propto \dfrac{1}{\lambda }$
Therefore photons of shortest wavelength will emit maximum energy. From the options given we can say $16eV$ is maximum.