What is the change in energy ΔE of the hydrogen atom as the electron makes the transition from the n=3 energy level to the n=1 energy level?Express your answer numerically in electron volts.

An

Z is the atomic number , i.e. the number of protons in the nucleus, and RE is the Rydberg energy.

RE = mₑ ∙e⁴ / (2∙ħ²) ≈ 13.6 eV  

(mₑ mass of electron, e electron charge, ħ reduced Planck constant)    

A)  

Atomic number of hydrogen is Z = 1, so

E = - RE/n²

The change in energy of the hydrogen atom si the difference between the initial and final energy of the electron:

ΔE = E₁ - E₃ = - RE/1² -  (-RE/3²) = RE∙((1/9 - 1) = -12.1 eV

That means the energy of the atom reduces by 12.1 eV

B)

The energy of the emitted  photon equals the energy loss of the atom:

Ep = - ΔE = 12.1 eV

Energy Ep and wavelength λ of a photon are related as [2]:

Ep = h∙c / λ

whee h is Planck constant and c speed of light

Hence

λ = h∙c / Ep  

= 4.136×10⁻¹⁵ eV∙s  2.998×10⁸ m/s / 12.1 eV

= 1.026×10⁻⁷ m/s  

= 102.6 nm

C)

Helium has atomic number Z = 2, so the energy of an electron at level in He⁺ is:

E = - RE ∙ 2²/n²  = - 4∙RE/n²

The energy required to lift the electron from level 2 to some higher level n is:

ΔE = E_n - E₂ = - (4∙RE/n²)  - (- 4∙RE/2²) = RE∙(1 - (4/n²))

This energy difference equals the energy of the photon:

Ep = RE∙(1 - (4/n²))

<=>

(4/n²) = 1 - (Ep/RE)  

=>

n² = ( 4 / (1 - (Ep/RE)) )  

=>

n = √( 4 / (1 - (Ep/RE)) )  

= √( 4 / (1 - (12.01eV/13..6eV)) )  

= 6swer:

Explanation:

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its a 5.5 n

glad that