Question

An electron in the n = 6 level emits a photon with a wavelength of 410.2 nm. to what energy level does the electron move?



a. n = 1



b. n = 2



c. n = 3



d. n = 4



e. n = 5

Answer 1

Answer:

b. n = 2

Explanation:

To find the energy level you use the following formula from the Bohr's model:

$\Delta E=-13.6(\frac{1}{n_2^2}-\frac{1}{n_1^2})$   (1)

n2: initial state

n1: final state

To find the final state n2, is necessary to calculate the change in energy. This is made by using the following formula:

$\Delta E=h\nu=h\frac{c}{\lambda}$

h: Planck's constant = 6.62*10^⁻34Js

By replacing you obtain:

$\Delta E=(6.62*10^{-34}Js)\frac{3*10^8m/s}{410.2*10^{-9}m}=4.841*10^{-19}J$

4.841*10^{-19}(6.242*10^{18}eV)=3.022eV (to use the formula (1) the unit of energy must be eV).

Finally, by doing n1 the subject of the formula (1) you obtain:

$\Delta E+\frac{13.6}{n_2^2}=\frac{13.6}{n_1^2}\\\\n_1=\sqrt{\frac{13.6}{\Delta E+\frac{13.6}{n_2^2}}}=\sqrt{\frac{13.6}{3.022+\frac{13.6}{6^2}}}=2$

hence, the electron moves to the n=2 level

Answer 2

The energy level in which the electron moves is : ( B )  n = 2

Given data :

wavelength of photon ( β ) = 410.2 nM

n1 = final state

n2 = Initial state = 6

To determine the energy level which the electron will move will apply the Bohr's model formula

ΔE = $-13.6 ( \frac{1}{n_{2} ^{2} } - \frac{1}{n_{1} ^{2} } )$ -------- ( 1 )

First step : Determine the change in energy using the relation below

ΔE =  $h \frac{c}{\beta }$   ----- ( 2 )            where ; $v = \frac{c}{\beta }$

c = 3 * 10⁸ m/s

h ( Planck's constant ) = 6.62 * 10⁻³⁴ Js

β = 410.2 * 10⁻⁹ m

Insert values into equation ( 2 )

ΔE = 4.841 * 10⁻¹⁹ J   ≈ 3.022 eV

Final step : Determine the energy level moved by the electron

Back to equation ( 1 ) make n₁ subject of the equation

n₁ = $\sqrt{\frac{13.6}{3.022 + \frac{13.6}{6^{2} } } }$    = 2

Hence we can conclude that the electron moves to n = 2

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