DrBob222
answered
That is E in J/atom
That x 6.02E23 = E in J/mol
Then convert to kJ.
1/4 = 1/n^2
1/25 = 1/5^2
Step-by-Step Bot
answered
E = - 13.6 eV / n^2
Where:
E is the energy of the electron transition in electron volts (eV)
n is the principal quantum number
First, we need to calculate the energy of the electron when it is in the initial state (n=5):
E1 = -13.6 eV / 5^2 = -13.6 eV / 25 = -0.544 eV
Next, we calculate the energy of the electron when it is in the final state (n=2):
E2 = -13.6 eV / 2^2 = -13.6 eV / 4 = -3.4 eV
Now, we can calculate the energy difference between the two states:
ΔE = E2 - E1 = (-3.4 eV) - (-0.544 eV) = -2.856 eV
Finally, we convert the energy difference from electron volts (eV) to kilojoules per mole (kJ/mol). The conversion factor is 1 eV = 96.485 kJ/mol:
ΔE = -2.856 eV x 96.485 kJ/mol/eV ≈ -275.89 kJ/mol
Therefore, approximately 275.89 kJ/mol of energy is released when an electron transitions from n=5 to n=2 in a hydrogen atom. Note that the negative sign indicates the release of energy.
Explain Bot
answered
ΔE = - R * (1/n²f - 1/n²i)
Where:
ΔE represents the change in energy
R is the Rydberg constant (2.179 × 10⁻¹⁸ J)
n₂ (final energy level) is 2
n₅ (initial energy level) is 5
First, let's convert the Rydberg constant (R) from joules (J) to kilojoules (kJ):
R = 2.179 × 10⁻¹⁸ J = 2.179 × 10⁻²⁵ kJ
Now, we can substitute the values into the equation and calculate ΔE:
ΔE = - (2.179 × 10⁻²⁵ kJ) * (1/2² - 1/5²)
= - (2.179 × 10⁻²⁵ kJ) * (1/4 - 1/25)
= - (2.179 × 10⁻²⁵ kJ) * (25/100 - 4/100)
= - (2.179 × 10⁻²⁵ kJ) * (21/100)
Calculating the result:
ΔE = - 4.5779 × 10⁻²⁶ kJ
Therefore, approximately 4.5779 × 10⁻²⁶ kJ of energy is released when an electron makes a transition from n=5 to n=2 in a hydrogen atom.
