Question

1. An electron in an atom absorbs a photon with an energy of 2.38 eV and jumps from the n = 2 to n = 4 energy level in the atom. Calculate the wavelength of the photon absorbed by the electron.
2. The electron in the n = 4 level then jumps down to the n = 3 level, emitting a photon with a wavelength of 1.66 ?m. Calculate the energy of the photon emitted.

3. Calculate the difference in energy between the n = 2 and the n = 3 energy levels in this atom.

4. Calculate the frequency of the photon emitted when the electron jumps from the n = 3 to the n = 2 energy level.

Answer 1

1. $1.0\cdot 10^{-6}m$

First of all, let's convert the energy of the absorbed photon into Joules:

$E=2.38 eV \cdot (1.6\cdot 10^{-19}J/eV)=1.98\cdot 10^{-19} J$

The energy of the photon can be rewritten as:

$E=\frac{hc}{\lambda}$

where

h is the Planck constant

c is the speed of light

$\lambda$ is the wavelength of the photon

Re-arranging the formula, we can solve to find the wavelength of the absorbed photon:

$\lambda=\frac{hc}{E}=\frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{1.98\cdot 10^{-19} J}=1.0\cdot 10^{-6}m$

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

$\lambda=1.66\cdot 10^{-6}m$

So the energy of the emitted photon is given by the formula used previously:

$E=\frac{hc}{\lambda}$

and using

$\lambda=1.66\cdot 10^{-6}m$

we find

$E=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{1.0\cdot 10^{-6}m}=1.99\cdot 10^{-19}J$

converting into electronvolts,

$E=\frac{1.99\cdot 10^{-19} J}{1.6\cdot 10^{-19} J/eV}=1.24eV$

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.