Calculate the photon energy (in joules) for the light emitted from the hydrogen atom if ninitial = 4 and nfinal = 2.

To solve this, we must first find the value of the wavelength using the formula:

1/ʎ = R [1/(nfinal)^2 – 1/(ninitial)^2]

where ʎ is wavelength and R is Rydberg’s constant = 10,973,731.6 m-1

1/ʎ = 10,973,731.6 m-1 [1/2^2 – 1/4^2]

1/ʎ = 2,057,574.675 m-1

ʎ = 4.86 x 10^-7 m

Then compute for energy using the equation:

E = hc / ʎ

where h is Plancks constant = 6.63 x 10^-34 J s, c is speed of light = 3 x 10^8 m/s

E = (6.63 x 10^-34 J s) (3 x 10^8 m/s) / 4.86 x 10^-7 m

E = 4.09 x 10^-19 J

it should be p1 v1 = p2 v2

the dependant variable is spilled water

the independent is the heat