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A ground state hydrogen atom absorbs a photon of light having a wavelength of 92.05 nm. It then gives off a photon having a wavelength of 1736 nm. What is the final state of the hydrogen atom?

Respuesta :

taskmasters
Absorbed photon energy
Ea = hc/λ.. (Planck's equation)
Ea = hc / 92.05^-9m 

Energy emitted
Ee = hc/ 1736^-9m 

Energy retained ..
∆E = Ea - Ee = hc(1/92.05^-9 - 1/1736^-9) 
∆E = (6.625^-34)(3.0^8) (1.028^7)
∆E = 2.04^-18 J 

Converting J to eV (1.60^-19 J/eV)
 ∆E = 2.04^-18 / 1.60^-19
∆E = 12.70 eV 

Ground state (n=1) energy for Hydrogen = - 13.60eV 

New energy state = (-13.60 + 12.70)eV = -0.85 eV 

Energy states for Hydrogen
En = - (13.60 / n²) 

n² = -13.60 / -0.85 = 16
n = 4
onyebuchinnaji

The final state of the hydrogen atom is ground state after the emission of the photons.

The given parameters;

  • wavelength of the photon absorbed, λ₁ = 92.05 nm = 92.05 x 10⁻⁹ m
  • wavelength of the photon emitted, λ₂ = 1736 nm = 1736 x 10⁻⁹ m

The change in the energy of the photon is calculated as follows;

ΔE = hf₂ - hf₁

[tex]\Delta E = h(f_2 - f_1)\\\\\Delta E = h (\frac{c}{\lambda _2} - \frac{c}{\lambda _1} )\\\\\Delta E = h c (\frac{1}{\lambda _2} - \frac{1}{\lambda _1} )\\\\\Delta E = (6.626 \times 10^{-34})\times (3\times 10^8)(\frac{1}{1736 \times 10^{-9}}\ -\ \frac{1}{92.05 \times 10^{-9}} )\\\\\Delta E = -2.05 \times 10^{-18} \ J[/tex]

The new energy level is calculated as follows using Borh's model;

[tex]\Delta E = \frac{-13.6 \ eV}{n^2} \\\\\n^2 = \frac{-13.6 \times 1.6\times 10^{-19}}{-2.05 \times 10^{-18}} \\\\n^2 = 1.0615\\\\n = \sqrt{1.0615} \\\\n = 1[/tex]

Thus, the final state of the hydrogen atom is ground state after the emission of the photons.

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