The excited state of a certain atom is 32 eV 021 eV h 6626

The excited state of a certain atom is 3.2 eV ± 0.21 eV. (h = 6.626 × 10-34 J s, 1eV = 1.60 × 10-19 J) (a) What is the average lifetime of this state, in femtoseconds? \' (b) If the excited energy were doubled (to 6.4 eV ± 0.21 eV), how would the lifetime be affected?

Solution

As given ,excited state of a certain atom is 3.2 eV ± 0.21 eV, we know that

In quantum mechanics, an excited state of a system (such as an atom,molecule or nucleus) is any quantum state of the system that has a higherenergy than the ground state (that is, more energy than the absolute minimum).

The lifetime of a system in an excited state is usually short: spontaneousor induced emission of a quantum of energy (such as a photon or aphonon) usually occurs shortly after the system is promoted to the excited state, returning the system to a state with lower energy (a less excited state or the ground state). This return to a lower energy level is often loosely described as decay and is the inverse of excitation.

Although findng carrier lifetime is a non trivial task, we can simply it to

Et

Hence, the life time will come to

t = 1.3 +- .2 fS

With increasing energy lifetime will decrease. In that case it iwll become half.