An electron in the n = 3 energy level of the hydrogen atom emits a photon with wavelength 656.27 nm. What is the change in energy of the electron, and to which energy level does it move?

change in energy of the electron=3.052x 10^-19J

which energy level does it move= level 2 , n=2

Explanation:

Using the formulae

1/λ = R (1/n1²- 1/n2²)

Where λ= 656.27 nm

1 nm = 1 x 10^-9 m

656.27 nm = 656.27 x 1 x 10^-9 =6.5726 x 10^-7

R =Rydberg constant = 1.0967 x 10^7m-1

1/λ = R (1/n1²- 1/n2²)

1/6.5726 x 10^-7=1.0967 x 10^7(1/n1²- 1/3²)

1/n1²=(1/6.5726 x 10^-7 x   1/1.0967 x 10^7) + 1/9

1/n1²=1,521,467.9 x 9.118x10^-8 + 0.1111

1/n1² =0.2498

n1²= 1/0.2498 =4

n1= = 2

it moves to energy level 2

b) Change in energy =ΔE = Rhc (1/n1²- 1/n2²)

Where R==Rydberg constant = 1.0967 x 10^7m-1

h = Planck constant = 6.626x 10^-34js

c = speed of light = 3.0 x 10^8 x m/s

ΔE = Rhc (1/n1²- 1/n2²)

=1.0967 x 10^7m-1 x6.626x 10^-34js X 3.0 x 10^8 x m/s (1/2² - 1/3²)

=2.18 x 10-18 x ( 1 /4 - 1/9)

=3.052x 10^-19J

n+h→nh3

we must remember that nitrogen and hydrogen are both diatomic molecules in their standard gas form. this adjusts the equation to

n2+h2→nh3

now we need to adjust coefficients in order to balance the atoms on each side of the equation. currently we have 2 atoms of nitrogen and 2 atoms of hydrogen on the reactant side and 1 atom of nitrogen and 3 atoms of hydrogen on the product side.

we can balance the hydrogens by placing a coefficient of 2 in front of ammonia and a coefficient of 3 in front of the hydrogen.

n2+3h2→2nh3

this gives us 6 hydrogen on each side and coincidentally the nitrogens now equal 2 on each side.

personal, she chose to do it.