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

i don't see a text, but i am gonna assume this *orbit* the curved path, usually elliptical, described by a planet, satellite, spaceship, etc., around a celestial body, as the sun.

the cold front replaces the warmer air at the ground level changing the wind to be breezy and denser.there is also an immediate drop in the temperature.