An electron in the 5th energy level of the H atom drops to the 3th energy level. In other words an electron in an excited state drops to a less excited state. What is the energy (in J) of the emitted photon

When an electron drops to a less excited state from a higher energy level, it gives off (or emits) energy in form of a photon. The energy can be calculated using Rydberg Formula:

[tex]E_{photon} = E_{o} (\frac{1}{n.final^{2} } - \frac{1}{ n.initial^{2}} )[/tex]

where:

n.final and n.initial are the initial and final energy levels

Eo is Rydberg's constant = 13.6eV ≡ 1.602 ×[tex]10^{-19}[/tex]J

[tex]E_{photon} = 1.602[/tex] × [tex]10^{-19}[/tex] [tex](\frac{1}{3^{2} } - \frac{1}{5^{2} } )[/tex]

∴ Energy emitted = 1.139 ×[tex]10^{-20} J[/tex]

aabdulsamir aabdulsamir · Answer 2 · 2020-03-24T01:53:25+01:00

Answer: 1.55 * 10-¹⁹ J

Explanation:

Using Rydberg's equation to calculate the wavelength of the transition of the photon from N = 5 to N = 3,

1/y = R [1/n²(final) - 1/n²(initial)]

y = wavelength (y is not the traditional sign of the Greek alphabet lambda, but however due to some restrictions currently, I'll be using letter "y" to represent lambda in my calculation)

R = Rydberg's constant = 1.0974 * 10-⁷ m-¹

1/y = 1.0974 * 10⁷ * [1/3³ - 1/5²]

1/y = 7.80 * 10⁵ m-¹

y = 1.28 * 10-⁶ m

Using energy-wavelength relationship,

E= hc/y

h = plank's constant

c = speed of light

y = wavelength

E = energy

E = (6.626 * 10-³⁴ * 3.0 * 10⁸) / 1.28 * 10-⁶

E = 1.55 * 10-¹⁹ J

Energy dissipated during the transition was 1.55 * 10-¹⁹ J