Find out the energy of emitted photon when an electron in Li2+ ion returns to its ground state from n=2 level.
energy in electron volts= 13.6 (3^2) (1/1 -1/2^2)
= .75*13.6*9 electron volts
To find the energy of the emitted photon when an electron in Li2+ ion returns to its ground state from the n=2 level, we can use the formula:
E = -RH(Z^2 / n^2)
Where:
E is the energy of the emitted photon,
RH is the Rydberg constant (2.18 x 10^-18 J),
Z is the atomic number of the nucleus (3 for Li2+),
and n is the principal quantum number (2 in this case).
Plugging in the values, we get:
E = -RH(3^2 / 2^2)
Simplifying further:
E = -RH(9 / 4)
Now, we can calculate the numerical value of the energy using the given value of the Rydberg constant:
E = -(2.18 x 10^-18 J)(9 / 4)
E ≈ -4.91 x 10^-18 J
Note that the negative sign indicates that energy is being released as a photon.
To find the energy of the emitted photon when an electron in Li2+ ion returns to its ground state from n=2 level, we can use the equation:
ΔE = E_final - E_initial
Where ΔE is the change in energy, E_final is the energy of the final state, and E_initial is the energy of the initial state.
The energy of an electron in the nth energy level of a hydrogen-like atom or ion can be given by:
En = -13.6 eV / n^2
For the Li2+ ion, the ground state is n=1 and the initial state is n=2. Therefore, the energy of the initial state (E_initial) can be calculated as:
E_initial = -13.6 eV / (2^2)
Now, we need to find the energy of the final state, which is the ground state (n=1). Plugging n=1 into the formula, we get:
E_final = -13.6 eV / (1^2)
Now we can calculate the change in energy (ΔE):
ΔE = E_final - E_initial
Substitute the values:
ΔE = (-13.6 eV / (1^2)) - (-13.6 eV / (2^2))
Simplify the expression:
ΔE = -13.6 eV - (-13.6 eV / 4)
ΔE = -13.6 eV + 3.4 eV
ΔE = -10.2 eV
So, the energy of the emitted photon when an electron in Li2+ ion returns to its ground state from n=2 level is 10.2 eV.