1. Interpret the process

A free electron with kinetic energy $2.6\,\text{eV}$ combines with a proton $\left(\text{H}^+\right)$ to form a hydrogen atom in the first excited state.

For hydrogen:

• Ground state: $n=1$, energy $=-13.6\,\text{eV}$
• First excited state: $n=2$, energy $=-3.4\,\text{eV}$

So after collision, the electron is captured into the $n=2$ state.

2. Energy released as photon

Initially:

• Proton is free
• Electron is free with kinetic energy $2.6\,\text{eV}$

Taking zero energy for a free electron-proton system, initial total energy is $E_i = 2.6\,\text{eV}.$

Finally, hydrogen atom is in $n=2$ state, so $E_f = -3.4\,\text{eV}.$

Hence photon energy released is $E_{\gamma} = E_i - E_f = 2.6 - (-3.4) = 6.0\,\text{eV}.$

3. Convert photon energy into joules

Using $1\,\text{eV} = 1.6\times 10^{-19}\,\text{J},$ we get $E_{\gamma} = 6.0\times 1.6\times 10^{-19} = 9.6\times 10^{-19}\,\text{J}.$

4. Use $E = h\nu$

Given $h = 6.6\times 10^{-34}\,\text{J s}$

Therefore, $\nu = \frac{E}{h} = \frac{9.6\times 10^{-19}}{6.6\times 10^{-34}}.$

$\nu \approx 1.45\times 10^{15}\,\text{Hz}.$

5. Convert to MHz

Since $1\,\text{MHz} = 10^6\,\text{Hz},$

$\nu = \frac{1.45\times 10^{15}}{10^6} = 1.45\times 10^9\,\text{MHz}.$

6. Match with options

The frequency is $\boxed{1.45\times 10^9\,\text{MHz}}$

So the correct option is:

$\boxed{\text{C}}$

7. Comparison with stored correct answer

Stored correct answer: $\text{C}$

Our derived answer is also $\text{C}$, so they agree.