Recall Planck's constant equals

Question

Recall Planck's constant equals 6.63 × 10⁻³⁴ J·s and the speed of light is 3.00 × 10⁸ m/s.

 

Calculate the wavelength (in nm) of a photon emitted by a hydrogen atom when its electron drops from the n = 5 state to the n = 4 state.

(Answer in nm)

Answer 1

EQUATIONS YOU NEED TO KNOW TO SOLVE THIS PROBLEM

1). E_n = R_H(1/n_f² - 1/n_i²) R_H = Rydberg's constant = (2.18 x 10^-18J) f = final state i = initial state

2). E = hc/λ h = Planck's constant c = speed of light

To Solve this problem you need to find the Energy from state 5 to state 4

n = 5 is initial state

n = 4 is final state

Using Equation 1 to find the energy

E = λ(2.18 x 10^-18J)(1/4² - 1/5²) = 4.91 x 10^-20

Using Equation 2 to find the wavelength

*if E = hc/λ then this equation can be rearranged to λ = hc/E

λ = ((6.63 x 10^-34J•s)(3.0 x 10⁸m/s))/(4.91 x 10^-20) = 4.05 x 10^-6m

Our answer is in meters and we need to convert to nm: 1m = 10⁹nm

(4.05 x 10^-6m)(10⁹nm) = 4050 nm

Answer 2

Hello, Rashed,

[Update: The energy I obtained for the electron transition is incorrect. It should be 4.6x10^-20 J, as per Kimberly M's comment and posted answer. Please use that number in the calculations.]

I need to assume the value of the energy change for the electron moving from n=5 to n=4 is accurate in a chart of hydrogen energy levels. I get a change of 1.9 eV, which is equal to 2.227x10^-19J. Please check this against a table or calculation provided in your text. [Incorrect value: should be 4.6x10^-20J]

Given that E = hc/λ, we can rearranged to find λ, the wavelength.

λ = hc/E

λ = (6.63 × 10⁻³⁴ J·s)(3.00 × 10⁸ m/s)/(2.227x10^-19J)

λ = 8.93x10^-7m

λ = (8.93x10^-7m)*(1.0x10⁹ nm/m) = 893 nm [Needs corrected energy value]

Please check my assumptions & calculations.

Bob