To determine the energy of the photon emitted during the combination of the maximum energy photoelectron with a negative particle, we first need to analyze the situation step by step. We know that the energy of the incident radiation and the work function of the surface play crucial roles in this process.

Understanding Photoelectron Emission

When 400 nm radiation strikes a surface with a work function of 1.9 eV, we can calculate the energy of the incoming photons using the formula:

E = h * c / λ

Where:

• E is the energy of the photon.
• h is Planck's constant (4.14 x 10^-15 eV·s).
• c is the speed of light (approximately 3 x 10⁸ m/s).
• λ is the wavelength of the radiation (400 nm or 400 x 10^-9 m).

Plugging in the values:

E = (4.14 x 10^-15 eV·s) * (3 x 10⁸ m/s) / (400 x 10^-9 m)

Calculating this gives:

E ≈ 3.1 eV

Calculating Maximum Energy of Photoelectrons

The maximum kinetic energy of the emitted photoelectrons can be found using the photoelectric equation:

K.E. = E - Work Function

Substituting the known values:

K.E. = 3.1 eV - 1.9 eV = 1.2 eV

Photon Emission During Combination

When the maximum energy electron (with 1.2 eV of kinetic energy) combines with a negative particle, energy is released in the form of a photon. The energy of the emitted photon can be calculated based on the energy levels of the resulting He⁺ ion.

Since the He⁺ ion is in its fourth excited state, we need to determine the energy difference between this state and the ground state. The energy levels of hydrogen-like atoms can be calculated using the formula:

E_n = -13.6 eV / n²

For the fourth excited state (n = 4):

E_4 = -13.6 eV / 4² = -13.6 eV / 16 = -0.85 eV

The ground state energy (n = 1) is:

E_1 = -13.6 eV

The energy difference (which corresponds to the photon emitted) when the ion transitions from n = 4 to n = 1 is:

ΔE = E_1 - E_4 = -13.6 eV - (-0.85 eV) = -13.6 eV + 0.85 eV = -12.75 eV

However, this value is not in the 2 to 4 eV range. We need to consider transitions between other energy levels as well. For example, the transition from n = 4 to n = 3 would yield:

E_3 = -13.6 eV / 3² = -13.6 eV / 9 ≈ -1.51 eV

Thus, the energy of the photon emitted during this transition would be:

ΔE = E_3 - E_4 = -1.51 eV - (-0.85 eV) = -1.51 eV + 0.85 eV = -0.66 eV

Continuing this process for transitions from n = 4 to n = 2:

E_2 = -13.6 eV / 2² = -13.6 eV / 4 = -3.4 eV

Then:

ΔE = E_2 - E_4 = -3.4 eV - (-0.85 eV) = -3.4 eV + 0.85 eV = -2.55 eV

Final Consideration

From the calculations, we see that the emitted photon energies from transitions between the excited states of He⁺ can indeed fall within the 2 to 4 eV range, specifically from transitions like n = 4 to n = 2. Therefore, the energy of the photon emitted during the combination can be in the range of approximately 2.55 eV, which is within the specified range of 2 to 4 eV.

In summary, the energy of the photon emitted during and after the combination of the maximum energy photoelectron with the negative particle can indeed lie within the desired range, depending on the specific transitions occurring in the He⁺ ion.