Explanation:
Wavelength in an emission spectrum, \(\lambda=435\ nm=435\times 10^{-9}\ m\)
The energy of an electron is given by :
\(E=\dfrac{hc}{\lambda}\)
Where
h is the Planck's constant
c is the speed of light
For 435 nm, the energy of the electron will be :
\(E=\dfrac{6.63\times 10^{-34}\times 3\times 10^8\ m/s}{435\times 10^{-9}}\)
\(E=4.57\times 10^{-19}\ J\)
We know that \(1\ eV=1.6\times 10^{-19}\ J\)
So, \(E=\dfrac{4.57\times 10^{-19}}{1.6\times 10^{-19}}\)
So, E = 2.86 eV
The energy of the electron dropping from one energy level is 2.86 eV. We know that,
\(\dfrac{hc}{\lambda}=E_{n_2}-E_{n_1}\)
From the given energy levels :
\(E_5-E_2=-0.544-(-3.403)\)
\(E_5-E_2=2.859\ eV\)
So, the transition must be from E₅ to E₂. Hence, this is the required solution.
