An electrom in the hydrogen atom makes a transition from
amenergy state of principal quantum numbers ni to the
n=2state. if the photon emitted has a wave length of 434nm, what
isthe value of ni?

ni = approx 1

Explanation:

speed of light c = fλ where is the the corresponding frequency and  λ is the wavelength.

λ = 434 nm, 1 nm =  1/ 10⁹ m

3.00 × 10⁸ m/s = (434 / 10 ⁹) m × f

f = 3.00 × 10⁸ × 10⁹) / 434 = 6.9124 × 10 ¹⁴ Hz

Energy = hf where h is planck constant = 6.62 × 10⁻³⁴ J.s

E = 6.62 × 10⁻³⁴ J.s × 6.9124 × 10 ¹⁴ Hz = 4.76 × 10⁻¹⁹J

In a hydrogen atom, the energy of the photon emitted can be calculated using the formula below:

ΔE = RH ( 1/ ni² - 1/nf²)

4.76 × 10⁻¹⁹J  = 2.18 × 10⁻¹⁸ J ( 1/ ni² - 1/nf²)

(4.76 × 10⁻¹⁹J) / (2.18 × 10⁻¹⁸ J) = ( 1/ ni² - 1/nf²)

0.21835 = 1/ni² - 1/ 2²

0.21835 + 0.25 = 1/ni²

0.46835 = 1/ni²

ni² = 1 / 0.46835 = 2.135155

ni = √2.135155 = 1.46  approx 1