Question

Calculate, according to the Bohr model, the speed of the electron in the ground state of the hydrogen atom.

Short answer

Answer: The speed of an electron in the ground state of a hydrogen atom, according to the Bohr model, is approximately 2.188 * 10^6 m/s.

Step-by-step solution

Remember the Bohr model formula for the speed of an electron in an atom

According to the Bohr model, the speed of an electron in an orbit is given by the formula: v = (Z * e^2) / (2 * epsilon_0 * h * n) where: v is the speed of the electron, Z is the atomic number of the element, e is the elementary charge (1.602 * 10^{-19} C), epsilon_0 is the vacuum permittivity (8.854 * 10^{-12} C^2/N m^2), h is the Planck's constant (6.626 * 10^{-34} Js), and n is the principal quantum number (energy level).

Set the values for hydrogen in the ground state

Since we are working with a hydrogen atom, its atomic number Z = 1. The ground state corresponds to the lowest energy level, so n = 1. Now we can plug in these values into the formula: v = (1 * 1.602 * 10^{-19}) / (2 * 8.854 * 10^{-12} * 6.626 * 10^{-34} * 1)

Calculate the speed of the electron

By calculating the above expression, we get: v = (1.602 * 10^{-19}) / (2 * 8.854 * 10^{-12} * 6.626 * 10^{-34}) v ≈ 2.188 * 10^6 m/s So, the speed of the electron in the ground state of the hydrogen atom, according to the Bohr model, is approximately 2.188 * 10^6 m/s.