Answer:
B E = 2.18 x 10⁻¹⁸ J
E E = - 2.04 x 10⁻¹⁸ J
Explanation:
According to Bohr´s model ofthe atom the energy change for a given electronic transition between energy levels can be determined using the Rydberg´s equation . Therefore, the strategy here is to use Rydberg´s equation:
1/λ = Rh x ( 1/n₁² - 1/n₂² )
where Rh is Rydberg´s constant ( 1.097 x 10⁷ / m ), n₁ and n₂ are the energy levels in the transition, and λ is the wavelegth of the transition.
Once 1/λ is determined, we can calculate the ionization energy using the relation: E = h c/λ = E hc (1/λ ) , where h is Planck´s constant, and c is the speed of light.
1/λ = Rh x ( 1/n₁² - 1/n₂² ) = 1.097 x 10⁷ / m x ( 1/1² )
The term 1/n₂² goes to zero as n₂ tends to infinity.
E = 6.626 x 10⁻³⁴ J·s x 3 x 10⁸ m/s x 1.097 x 10⁷ /m
E = 2.18 x 10⁻¹⁸ J
For the second part we use the same equation but with n₁ = 1 and n₂ = 4
1/λ = Rh x ( 1/n₁² - 1/n₂² ) = 1.097 x 10⁷ /m x ( 1/1² -1/4² )
1/λ = 1.03 x 10⁷/m
E = -6.626 x 10⁻³⁴ J·s x 3 x 10⁸ m/s x 1.03 x 10⁷ /m
E = - 2.04 x 10⁻¹⁸ J
( When using Rydbergs equation by convention n₁ is the lowest energy level and the sign will always will come positive, but since here we are talking in going from level 4 to level 1, energy will be released hence the negative sign )
