Question

Using the Rydberg equation, calculate the energy in joules, J, of the emission line when electrons drop from n₂ = 4 to n₁= 2 in the hydrogen atom.

Hint: Answer includes 3 significant figures, use the format X.XXEX for scientific notation. Relevant constants: h = 6.626 x 10^-34 J·s, c = 3.00 x 10⁸ m·s^-1, and R_H = 1.097 x 10⁷ m^-1.

Alright I tried to plug in I am getting 1.36x10^-18 what am I doing wrong?

is this right 0.0005 or 0.000486174

Answer 1

1 /    = R_H [1/ n₁² - 1/n₂²]

            = 1.0973 x 10^7 [1/2^2 - 1 / 4^2]

           = 2.06 x 10^6 m-1

1 / = 2.06 x 10^6 m-1

E = h C /

   = 6.626 x 10^-34 x 3.00 x 10⁸ x 2.06 x 10^6

   = 4.09 x 10^-19 J

Energy = 4.09 x 10^-19 J