Calculate the wavelength of radiation which excites the electrons of hydrogen atom from ground state to fourth energy level.ionization energy of hydrogen is 1312kj/mol
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Energy of nth orbit is given by En = -21.8 × 10⁻¹⁹ × Z²/n²
If Z = 1 then, En = -21.8 × 10⁻¹⁹/n² J/atom
For fist orbit , E₁ = -2.18 × 10⁻¹⁹/1² = -21.8 × 10⁻¹⁹ j/atom
for 4th orbit , E₄ = -2.18 × 10⁻¹⁹/4² = -1.36 × 10⁻¹⁹ J/atom
Now, ∆E = E₄ - E₁ = hc/λ
-1.36 × 10⁻¹⁹ + 2.18 × 10⁻¹⁹ = 6.636 × 10⁻³⁴ × 3 × 10⁸/λ
2.04 × 10⁻¹⁹ = 6.636 × 3 × 10⁻²⁶/λ
λ = 97.3 nm
If Z = 1 then, En = -21.8 × 10⁻¹⁹/n² J/atom
For fist orbit , E₁ = -2.18 × 10⁻¹⁹/1² = -21.8 × 10⁻¹⁹ j/atom
for 4th orbit , E₄ = -2.18 × 10⁻¹⁹/4² = -1.36 × 10⁻¹⁹ J/atom
Now, ∆E = E₄ - E₁ = hc/λ
-1.36 × 10⁻¹⁹ + 2.18 × 10⁻¹⁹ = 6.636 × 10⁻³⁴ × 3 × 10⁸/λ
2.04 × 10⁻¹⁹ = 6.636 × 3 × 10⁻²⁶/λ
λ = 97.3 nm
Answered by
1
Explanation:
Ionization energy of hydrogen atom=−13.6Z
2
=−13.6eV
E
ionization
=
λ
hc
λ
ionisation
6.626×10
−34
×3×10
8
=13.6×1.6×10
−19
λ
ionisation
=912×10
−8
m
ionization energy for 1 mole H=E
ionization
×N
A
=13.6×1.6×10
−19
×6.023×10
23
=1310kJ/mole
=1313kJ/mol(approx)
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