Answer:
E = 3.6 x 10⁶ J/mol
f = 1.1 x 10 ¹⁶ s⁻¹
λ = 2.6 x 10⁻⁸ m
Explanation:
Rydberg´s equation for hydrogen-like atoms is:
1/λ = Z²Rh (1/n₁² - 1/n₂²)
where λ = wavelength
Z² = atomic number of hydrogen-like atom
Rh= Rydberg´s constatn
n₁ = principal quantum number of initial state
n₂ = principal quantum number of final state
We also know that E = h(c/ λ ) = hf, where f is frequency equal to c/λ, so we have all the information needed to answer the questions.
a) We are asked the energy to remove the electron from 1 mol of B⁴⁺ , that means the transition is from n₁ = 3 to n₂ = ∞. The term 1/n₂ approaches zero in the infinity so:
Working in SI units
1/λ = 5² x1.097 x 10⁷ m⁻¹ ( 1/3² - 0) = 3.0 x 10⁷ m⁻¹
E= h(c/ λ )= hc(1/ λ) = 6.626 x 10⁻³⁴ J/s x 3 x 10⁸ m/s x (3.0 x 10⁷ m⁻¹)
= 6.0x 10⁻¹⁸ J
This is the energy per atom, so per mol of atoms is
= 6.0x 10⁻¹⁸ J/atom x 6.022 x 10²³ atoms/mol = 3.6 x 10⁶ J/mol
b) f and λ from a transition n= 3 to n=2
1/ λ = 5² x1.097 x 10⁷ m⁻¹ x ( 1/2² - 1/3²) = 3.8 x 10⁷ m⁻¹ ⇒
λ = 1/ 3.8 x 10⁷ m⁻¹ = 2.6 x 10⁻⁸ m
f = 3 x 10⁸ m/s / 2.6 x 10⁻⁸ m = 1.1x 10 ¹⁶ s⁻¹
