Use the bohr model to calculate the radius and the energy of the b⁴⁺ ion in the n 3 state. how much energy would be required to remove the electrons from 1 mol of b⁴⁺ in this state? what frequency and wavelength of light would be emitted in a transition from the n 3 to the n 2 state of this ion? express all results in si units.

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

since the atomic number is the number of protons in the nucleus, as we move to the left on the periodic table, the electrostatic pull between the nucleus and the valence electron shell decreases. as we move down a group (vertical column), the valence shell energy level gets further and further from the nucleus.

1.43 x 10^-11 j the guy commented the awnser just making sure u see it .