1)Five possible transitions for a hydrogen atom are listed below: Select whether the atom gains or loses energy for each transition.

a.
Loses Gains ni = 6; nf = 4
Loses Gains ni = 4; nf = 2
Loses Gains ni = 5; nf = 3
Loses Gains ni = 2; nf = 5
Loses Gains ni = 5; nf = 7
b. Find the transition where the atom gains the most energy. How much energy does the atom gain?
c. Find the transition which will emit the shortest wavelength photon. What is the energy of the photon of this transition

1) The atom gains energy if nf > ni

"i" signifies "initial" and "f" signified "final".

2) For a hydrogenic one-electron aton, the atom gains the most energy when
1/ni^2 - 1/nf^2 is largest.

3) To emit a photon, the atom must lose energy. For the shortest wavelength,
1/ni^2 -1/nf^2 is the most negative.

To compute the energy gain use the Rydberg relationship. If you don't know it, check it out at
http://en.wikipedia.org/wiki/Rydberg_formula

Thank you for your timely help.

a.

Loses ni = 6; nf = 4 - Loses energy
Gains ni = 4; nf = 2 - Gains energy
Loses ni = 5; nf = 3 - Loses energy
Gains ni = 2; nf = 5 - Gains energy
Loses ni = 5; nf = 7 - Loses energy

b. To find the transition where the atom gains the most energy, we need to compare the differences in energy levels between the initial (ni) and final (nf) states of each transition.

The transition with the largest energy gain is when ni = 4 and nf = 2.

To calculate the energy gained, we can use the formula:

ΔE = 13.6 eV * (1/nf² - 1/ni²)

Substituting ni = 4 and nf = 2 into the formula:

ΔE = 13.6 eV * (1/2² - 1/4²)
ΔE = 13.6 eV * (1/4 - 1/16)
ΔE = 13.6 eV * (3/16)
ΔE ≈ 2.55 eV

Therefore, the atom gains approximately 2.55 electron volts (eV) of energy.

c. The shortest wavelength corresponds to the transition with the highest energy difference.

For this, we need to find the transition with the highest difference between initial and final energy levels. Looking at the provided transitions, the transition with ni = 2 and nf = 5 has the highest energy difference.

To calculate the energy of the photon emitted during this transition, we can use the formula:

E = 13.6 eV * (1/nf² - 1/ni²)

Substituting ni = 2 and nf = 5 into the formula:

E = 13.6 eV * (1/5² - 1/2²)
E = 13.6 eV * (1/25 - 1/4)
E = 13.6 eV * (4/100 - 25/100)
E = 13.6 eV * (-21/100)
E ≈ -2.856 eV

Since the energy of a photon emitted during a transition needs to be positive, we consider the magnitude. Thus, the energy of the emitted photon is approximately 2.856 eV.

a. To determine whether the atom gains or loses energy for each transition, we can use the equation:

ΔE = E_final - E_initial

Where ΔE represents the change in energy, E_final is the energy of the final state, and E_initial is the energy of the initial state.

For each transition:

- ni = 6; nf = 4 : The atom loses energy because the final state has a lower energy level than the initial state.
- ni = 4; nf = 2 : The atom loses energy.
- ni = 5; nf = 3 : The atom loses energy.
- ni = 2; nf = 5 : The atom gains energy because the final state has a higher energy level than the initial state.
- ni = 5; nf = 7 : The atom gains energy.

b. To find the transition where the atom gains the most energy, we need to compare the energies of the initial and final states for each transition. Specifically, we want to find the transition with the largest difference in energy.

From the given transitions, the transition with the largest energy difference (gains the most energy) is ni = 2; nf = 5. To calculate the exact energy gained, we need to know the energy levels corresponding to ni and nf. In this case, the initial energy level (ni = 2) corresponds to the Lyman series, and the final energy level (nf = 5) corresponds to the Paschen series.

The equation to calculate the energy of a transition in the hydrogen atom is given by:

E = -13.6 eV * (Z^2 / n^2)

Where Z is the atomic number (1 for hydrogen) and n is the principal quantum number.

For ni = 2 (Lyman series):
E_initial = -13.6 eV * (1^2 / 2^2) = -13.6 eV * 0.25 = -3.4 eV

For nf = 5 (Paschen series):
E_final = -13.6 eV * (1^2 / 5^2) = -13.6 eV * 0.04 = -0.544 eV

Therefore, the atom gains energy:
ΔE = E_final - E_initial = (-0.544 eV) - (-3.4 eV) = 2.856 eV

c. To find the transition that will emit the shortest wavelength photon, we need to determine the corresponding energy levels and calculate the energy difference.

The energy of a photon is given by the equation:

E = hc / λ

Where E is the energy of the photon, h is Planck's constant (6.62607015 × 10^-34 J·s), c is the speed of light (2.998 × 10^8 m/s), and λ is the wavelength of the photon.

Since we want the shortest wavelength, we are looking for the transition with the highest energy difference.

From the given transitions, the transition that corresponds to the largest energy difference is ni = 6; nf = 4. To calculate the energy difference, we can use the same equation as in part b:

E_initial = -13.6 eV * (1^2 / 6^2) = -13.6 eV * 0.0278 = -0.377 eV
E_final = -13.6 eV * (1^2 / 4^2) = -13.6 eV * 0.0625 = -0.85 eV

ΔE = E_final - E_initial = (-0.85 eV) - (-0.377 eV) = -0.473 eV

Finally, we can calculate the corresponding wavelength using the energy of the photon:

E = -0.473 eV = (-0.473 eV) * (1.602 × 10^-19 J/eV) = -7.57 × 10^-20 J
λ = hc / E = (6.62607015 × 10^-34 J·s * 2.998 × 10^8 m/s) / (-7.57 × 10^-20 J)
λ ≈ 82.77 nm

Therefore, the transition from ni = 6 to nf = 4 will emit a photon with a wavelength of approximately 82.77 nm.