It is possible to determine the ionization energy for hydrogen using the Bohr equation. Calculate the ionization energy for an atom of hydrogen, making the assumption that ionization is the transition from n=1 to n=infinity.

A. - 2.18 x 10-18 J

B. + 2.18 x 10-18 J

C. + 4.59 x 10-18 J

D. - 4.59 x 10-18 J

E. + 4.36 x 10-18 J

Calculate the energy change associated with the transition from n=4 to n=1 in the hydrogen atom.

A. + 4.89 x 10-18 J

B. + 1.64 x 10-18 J

C. - 6.12 x 10-18 J

D. + 3.55 x 10-18 J

E. - 2.04 x 10-18 J

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)