An 8.00 -eV photon is absorbed by an electron in the \(n=2\) state of a hydrogen atom. Calculate the final speed of the electron.

We can determine the electron's initial energy using the formula for energy levels in a hydrogen atom: $$E_n = -\frac{13.6 \,\text{eV}}{n^2}$$ where \(n\) is the energy level number. In this case, \(n=2\). $$E_2 = -\frac{13.6 \,\text{eV}}{2^2} = -3.4 \,\text{eV}$$

When the electron absorbs the 8.00 eV photon, its energy increases by the photon's energy: $$E_{total} = E_2 + E_{photon} = -3.4 \,\text{eV} + 8.00 \,\text{eV} = 4.6 \,\text{eV}$$

If the new energy state is in the nth energy level, we have: $$E_n=-\frac{13.6 \,\text{eV}}{n^2} = 4.6 \,\text{eV}$$ Solving for n: $$n^2 = \frac{-13.6 \,\text{eV}}{4.6 \,\text{eV}}$$ However, we find that the resulting \(n^2\) is negative which means that the electron is no longer bound to the atom, and it's in a continuous energy state or free electron.

In this case, since the electron is in a continuous energy state, its total energy is equal to its kinetic energy: $$E_{f,k} = E_{total} = 4.6 \,\text{eV}$$

Now we can calculate the speed of the electron using the kinetic energy formula: $$E_{f,k} = \frac{1}{2} mv^2$$ Where \(m\) is the electron mass, and \(v\) is the final speed. First, we need to convert the kinetic energy from eV to Joules: $$E_{f,k} = 4.6 \,\text{eV} \cdot \frac{1.6 \times 10^{-19} \,\text{J}}{\text{eV}} = 7.36 \times 10^{-19} \,\text{J}$$ Now, we can solve for the final speed: $$v = \sqrt{\frac{2E_{f,k}}{m}}$$ Taking \(m\) as the mass of an electron \(9.11 \times 10^{-31} \,\text{kg}\): $$v = \sqrt{\frac{2(7.36 \times 10^{-19} \,\text{J})}{9.11 \times 10^{-31} \,\text{kg}}} = 1.27 \times 10^6 \,\text{m/s}$$ The final speed of the electron after absorbing the 8.00 eV photon is approximately \(1.27 \times 10^6 \, \text{m/s}\).