A hydrogen atom is in an excited state with n = 2. Using the Bohr model of hydrogen, find the following.the radius of the electron orbit (in nm).Answer for part 1the angular momentum of the electron - use SI unitsAnswer for part 2the kinetic energy of the electron (in eV).Answer for part 3the potential energy of the electron (in eV).  Note that this may be a negative number.Answer for part 4the total energy of the electron (in eV). Note that this may be a negative number.

Part 1: The radius of the electron orbit

The radius of the electron orbit in a hydrogen atom according to the Bohr model is given by the formula:

r = n² * r₀

where r₀ is the Bohr radius (approximately 0.529 Å or 0.0529 nm), and n is the principal quantum number. In this case, n = 2.

So, r = 2² * 0.0529 nm = 0.2116 nm

Part 2: The angular momentum of the electron

The angular momentum (L) of the electron in a hydrogen atom according to the Bohr model is given by the formula:

L = n * h / 2π

where h is Planck's constant (approximately 6.626 x 10^-34 Js), and n is the principal quantum number. In this case, n = 2.

So, L = 2 * 6.626 x 10^-34 Js / 2π = 2.11 x 10^-34 Js

Part 3: The kinetic energy of the electron

The kinetic energy (K) of the electron in a hydrogen atom according to the Bohr model is given by the formula:

K = 13.6 eV / n²

where n is the principal quantum number. In this case, n = 2.

So, K = 13.6 eV / 2² = 3.4 eV

Part 4: The potential energy of the electron

The potential energy (U) of the electron in a hydrogen atom according to the Bohr model is given by the formula:

U = -2 * K

So, U = -2 * 3.4 eV = -6.8 eV

Part 5: The total energy of the electron

The total energy (E) of the electron in a hydrogen atom according to the Bohr model is given by the formula:

E = K + U

So, E = 3.4 eV + (-6.8 eV) = -3.4 eV