A free electron of 2.6 eV energy collides with a H$$^+$$ ion. This results in the formation of a hydrogen atom in the first excited state and a photon is released. Find the frequency of the emitted photon. (h = $$6.6 \times 10^{-34}$$ J s)

We begin with the given data. The incident free electron possesses a kinetic energy of $$2.6 \text{ eV}$$, while a bare proton (H$$^+$$) has zero bound-state energy because the electron is initially infinitely far away. After the collision the electron is captured and the system becomes a hydrogen atom in its first excited state, namely the level with principal quantum number $$n = 2$$.

For a hydrogen atom the stationary-state energies are obtained from the well-known Bohr formula

$$E_n \;=\; -\dfrac{13.6\ \text{eV}}{n^2}\;.$$

Putting $$n = 2$$ gives the final bound energy of the electron as

$$E_{\,\text{final}} \;=\; -\dfrac{13.6\ \text{eV}}{2^2} \;=\; -\dfrac{13.6\ \text{eV}}{4} \;=\; -3.4\ \text{eV}.$$

Now we compute the total energy before and after the capture:

• Initial energy (electron free): $$E_{\,\text{initial}}\;=\; +2.6\ \text{eV}.$$ • Final energy (electron bound in $$n=2$$): $$E_{\,\text{final}}\;=\;-3.4\ \text{eV}.$$

The energy released is the difference

$$\Delta E \;=\; E_{\,\text{final}} \;-\; E_{\,\text{initial}}$$

$$\qquad\;=\;(-3.4\ \text{eV})\;-\;(+2.6\ \text{eV}) \;=\; -6.0\ \text{eV}.$$

The negative sign indicates that $$6.0\ \text{eV}$$ of energy is emitted as a photon. Hence the photon energy is

$$E_{\gamma} \;=\; 6.0\ \text{eV}.$$

To use Planck’s relation $$E = h\nu$$ we convert the energy into joules. Using $$1\ \text{eV} = 1.6 \times 10^{-19}\ \text{J}$$, we have

$$E_{\gamma} = 6.0\ \text{eV} \times 1.6 \times 10^{-19}\ \dfrac{\text{J}}{\text{eV}} = 9.6 \times 10^{-19}\ \text{J}.$$

Planck’s constant is given as $$h = 6.6 \times 10^{-34}\ \text{J s}.$$ Applying the formula $$\nu = \dfrac{E}{h}$$ gives

$$\nu \;=\;\dfrac{9.6 \times 10^{-19}\ \text{J}}{6.6 \times 10^{-34}\ \text{J s}}$$

$$\quad=\;\dfrac{9.6}{6.6}\times10^{(-19+34)}\ \text{s}^{-1}$$

$$\quad=\;1.4545 \times 10^{15}\ \text{Hz}.$$

Because the answer choices are expressed in megahertz (MHz), we convert Hertz to MHz using $$1\ \text{MHz} = 10^{6}\ \text{Hz}$$:

$$\nu \;=\;1.4545 \times 10^{15}\ \text{Hz} \;=\;1.4545 \times 10^{15} \div 10^{6}\ \text{MHz} \;=\;1.4545 \times 10^{9}\ \text{MHz}.$$

Rounding to three significant figures matches the option $$1.45 \times 10^{9}\ \text{MHz}$$.

Hence, the correct answer is Option B.