Calculate the energy difference for the transition of n=6 to n=2 for 1.00 mol of hydrogen atom?
To calculate the energy difference for the transition of an electron from one energy level to another in the hydrogen atom, we can use the energy formula derived from the Bohr model:
E = - 13.6 eV * (Z^2 / n^2)
Where:
E is the energy of the electron transition
Z is the atomic number (for hydrogen, Z = 1)
n is the principal quantum number of the initial or final energy level
To calculate the energy difference between two energy levels, we subtract the final energy from the initial energy:
ΔE = | Ef - Ei |
For the given transition from n=6 to n=2, we can calculate the energy difference as follows:
Initial energy (Ei) for n = 6:
Ei = - 13.6 eV * (1^2 / 6^2)
Final energy (Ef) for n = 2:
Ef = - 13.6 eV * (1^2 / 2^2)
Now, let's calculate the energy difference (ΔE):
ΔE = | Ef - Ei |
= | (-13.6 eV * (1^2 / 2^2)) - (-13.6 eV * (1^2 / 6^2)) |
Note: The negative sign in front of the energy values indicates that the electron is transitioning to a lower energy level.
To get the value of ΔE in electron volts (eV), we can perform the calculations:
ΔE = | (-13.6 eV * (1/4)) - (-13.6 eV * (1/36)) |
= | (-3.4 eV) - (-0.3778 eV) |
= | -3.4 eV + 0.3778 eV |
= | -3.0222 eV |
Finally, taking the absolute value of the energy difference, we get:
ΔE = 3.0222 eV
Hence, the energy difference for the transition of n=6 to n=2 for 1.00 mol of hydrogen atom is 3.0222 electron volts (eV).