1. [tex]1.0\cdot 10^{-6}m[/tex]
First of all, let's convert the energy of the absorbed photon into Joules:
[tex]E=2.38 eV \cdot (1.6\cdot 10^{-19}J/eV)=1.98\cdot 10^{-19} J[/tex]
The energy of the photon can be rewritten as:
[tex]E=\frac{hc}{\lambda}[/tex]
where
h is the Planck constant
c is the speed of light
[tex]\lambda[/tex] is the wavelength of the photon
Re-arranging the formula, we can solve to find the wavelength of the absorbed photon:
[tex]\lambda=\frac{hc}{E}=\frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{1.98\cdot 10^{-19} J}=1.0\cdot 10^{-6}m[/tex]
2. 1.24 eV
In this case, when the electron jumps from the n=4 level to the n=3 level, emits a photon with wavelength
[tex]\lambda=1.66\cdot 10^{-6}m[/tex]
So the energy of the emitted photon is given by the formula used previously:
[tex]E=\frac{hc}{\lambda}[/tex]
and using
[tex]\lambda=1.66\cdot 10^{-6}m[/tex]
we find
[tex]E=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{1.0\cdot 10^{-6}m}=1.99\cdot 10^{-19}J[/tex]
converting into electronvolts,
[tex]E=\frac{1.99\cdot 10^{-19} J}{1.6\cdot 10^{-19} J/eV}=1.24eV[/tex]
EDIT: an issue in Brainly does not allow me to add the last 2 parts of the solution - I have added them as an attachment to this post, check the figure in attachment.
