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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?

2 Answers

4 votes

Answer:

n = 1 :))))))))))))))))))

User Moshtaf
by
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5 votes

Answer:

ni = approx 1

Step-by-step 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

User Sparko Sol
by
5.0k points