B

Step 1: Analyze the initial conditions.
We have an electron with an initial energy of 20 eV. In joules, this energy can be calculated using the conversion factor where 1 eV = 1.6 × 10^-19 J:

$$ E_e = 20 ext{ eV} imes 1.6 imes 10^{-19} ext{ J/eV} = 3.2 imes 10^{-18} ext{ J} $$

Step 2: Determine the energy emitted by the hydrogen atom.
When the hydrogen atom transitions from an excited state back to the ground state, it emits radiation with a wavelength of 1.216 × 10^-7 m. The energy of the emitted photon can be calculated using the formula:

$$ E_{photon} = \frac{hc}{\lambda} $$

where h is Planck's constant (6.63 × 10^-34 J·s) and c is the speed of light (3 × 10⁸ m/s). Substituting the values:

$$ E_{photon} = \frac{(6.63 \times 10^{-34} ext{ J·s})(3 \times 10^{8} ext{ m/s})}{1.216 \times 10^{-7} ext{ m}} $$

This simplifies to:

$$ E_{photon} = \frac{1.989 \times 10^{-25} ext{ J·m}}{1.216 \times 10^{-7} ext{ m}} \approx 1.636 \times 10^{-18} ext{ J} $$

Step 3: Calculate the energy of the scattered electron.
The energy of the scattered electron can be determined by subtracting the energy of the emitted photon from the initial energy of the incoming electron:

$$ E_{e,scattered} = E_e - E_{photon} $$

Substituting in the values:

$$ E_{e,scattered} = 3.2 \times 10^{-18} ext{ J} - 1.636 \times 10^{-18} ext{ J} \approx 1.564 \times 10^{-18} ext{ J} $$

Step 4: Calculate the velocity of the scattered electron.
The kinetic energy of the scattered electron can be expressed using the formula:

$$ E_{k} = \frac{1}{2}mv^2 $$

where m is the mass of the electron (9.11 × 10^-31 kg). We will solve for v:

$$ v = \sqrt{\frac{2E_{k}}{m}} $$

Now substituting the values:

$$ v = \sqrt{\frac{2 \times 1.564 \times 10^{-18} ext{ J}}{9.11 \times 10^{-31} ext{ kg}}} $$

Performing the calculations:

$$ v = \sqrt{\frac{3.128 \times 10^{-18}}{9.11 \times 10^{-31}}} \approx \sqrt{3.43 \times 10^{12}} \approx 1.85 \times 10^{6} ext{ m/s} $$

Therefore, the velocity of the scattered electron is approximately 1.85 × 10⁶ m/s.