Answer to Question #143312 in General Chemistry for saan

Question

An electron in a hydrogen atom decays from a certain excited state down to the ground state. If the photon emitted has a wavelength of 97 nm, what was the n-value for the excited state?
E = hν = hc/λ
ΔE = RH(1/ni2 - 1/nf2)
RH = 2.18 x 10-18 J
Planck's constant = 6.63 x 10-34 J s
c = 3.00 x 108 m/s
1 nm = 10-9 m

Answer 1

Note that energy will be negative since its losing an atom

ΔE=(hc)/(v×RH)= [ 1/ni² - 1/nf² ]

-(6 63×10^-34 × 3.0×10⁸ )/(97×10^-9 ×2.18×10^-18) = (1/1² - 1/nf²)

-0.94=1-(1/nf²)

-1.94= -1/nf²

nf² = 1/1.94 = 0.515

nf = square root 0.515

nf= 0.72