Final answer:
A hydrogen atom in the ground state absorbs 10.2 eV of energy to transition to an excited state, resulting in an increase in the orbital angular momentum of the electron. The specific value of this increase requires quantum mechanical calculations and depends on the change in the principal quantum number from n=1 to a higher value.
Step-by-step explanation:
When a hydrogen atom in the ground state absorbs energy, it transitions to an excited state. Specifically, when 10.2 eV of energy is absorbed, the electron in a hydrogen atom transitions from the ground state to a higher energy level. This raises the orbital angular momentum of the electron, which is quantified by quantized values determined by the principal quantum number n. As the electron transitions to a higher orbital (n=1 to n=2 in this case), this is associated with specific quantized angular momentum changes.
The increase in orbital angular momentum for the electron involves a transition from one energy level to another, which is associated with a specific change in angular momentum according to the quantized nature of orbitals in a hydrogen atom, as determined by Bohr's model of the hydrogen atom and quantum mechanics. Each energy level of the hydrogen atom corresponds to a different angular momentum, with the first excited state being just one possible higher energy state.
The ground state orbital angular momentum is zero because n=1 yields an angular momentum of 0 due to the equation L = √(l(l+1))h/(2π), where l is the quantum number corresponding to the orbital angular momentum and is 0 in the ground state. Therefore, a transition to n=2 increases the angular momentum by the specific value mentioned in the question. However, the correct value for such an increase is not provided here as it would require additional calculation using quantum mechanics equations.