Yes, just multiply each phi by exp{-iEt/hbar} and add.Lee said:Oh, sorry.
phi1=Aexp(-r/ao)
phi2=B(1-r/ao)e(-r/2ao)
where ao is the Bohr radius.
As given by the question.
So can I take the Energy of each wave function (that I have already calculated) and apply the above substitution?
A wave function is a mathematical representation of the state of a quantum mechanical system, such as an electron. It describes the probability of finding the system in a particular state at a specific time.
The wave function of an electron tells us about its position, momentum, energy, and other physical properties. It also provides information about the probability of finding the electron in a particular location or state.
The wave function of an electron is determined by solving the Schrödinger equation, which is a fundamental equation in quantum mechanics. This equation takes into account the electron's energy, potential energy, and other physical properties to describe its behavior.
No, the wave function of an electron cannot be directly observed. It is a mathematical concept used to describe the behavior of a quantum mechanical system. However, the effects of the wave function can be observed through experiments and measurements.
The wave function of an electron changes over time according to the laws of quantum mechanics. It can be affected by external factors such as electric and magnetic fields, as well as interactions with other particles. The change in the wave function can also be described by the Schrödinger equation.