Question

Using the conversion factor from eV (electron volts) to joules, determine which energy line for an electron dropping from one energy level to another would show a line at 435 nm in an emission spectrum. Use the formulas and the energy level information below to answer the question.
Needed constants:

1.00eV = 1.6 * 10^-19 J
c = 3.00 * 10^8 m/s
h = 6.63 * 10^-34 J*s

Energy Level Values:
E6: E= -0.378 eV
E5: E= -0.544 eV
E4: E= -0.850 eV
E3: E= -1.51 eV
E2: E= -3.403V

Answer 1

The formula used to solve this is:
E = hc/λ

The given values can then be subsituted into the equation to sovle for the energy:

E = (6.63 x 10^-34 J-s) (3 x 10^8 m/s) / 435 x10^-9 m
E = 4.5724 x 10^-19 J

converting to electron volts
E = 4.5724 x 10^-19 J * 1.00eV / 1.6 * 10^-19 J
E = 2.86 eV

The closest energy line is
E2

Answer 2

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.