Step-by-step explanation:
According to Bohr's model, the electron in a hydrogen atom is in a circular orbit around the nucleus, and its angular momentum is quantized in integer multiples of Planck's constant h. The radius of the ground-state orbit is given by:
r1 = a0 = (4πε0ħ^2)/(me^2)
where ε0 is the permittivity of vacuum, me is the mass of the electron, and e is the elementary charge.
(a) The orbital speed of the electron can be computed as:
v1 = ħ/(m*r1)
where m is the mass of the electron. Substituting the values, we get:
v1 = (ħe^2)/(4πε0ħ^2m) = (e^2)/(4πε0ħ*m)
Plugging in the numerical values for the constants and mass, we get:
v1 = (9.0 x 10^9 m/s)
Therefore, the orbital speed of the electron in the ground state of hydrogen is approximately 9.0 x 10^9 m/s.
(b) The kinetic energy of the electron can be computed as:
KE1 = (1/2)mv1^2
Substituting the values, we get:
KE1 = (1/2)me*(e^2)/(4πε0ħ*m)^2
Plugging in the numerical values for the constants and mass, we get:
KE1 = (2.2 x 10^-18 J) = (13.6 eV)
Therefore, the kinetic energy of the electron in the ground state of hydrogen is approximately 13.6 electronvolts (eV).
(c) The electrical potential energy of the atom can be computed as:
PE1 = - (1/4πε0)*(e^2)/r1
Substituting the value of r1, we get:
PE1 = - (me^4)/(8ε0^2ħ^2)
Plugging in the numerical values for the constants and mass, we get:
PE1 = - (2.2 x 10^-18 J) = - (13.6 eV)
Therefore, the electrical potential energy of the ground state of hydrogen is approximately -13.6 eV. Note that the negative sign indicates that the electron is bound to the nucleus and that energy is required to remove it from the atom.