Question

A photon collides with a stationary hydrogen atom in the ground state inelastically. The energy of the colliding photon is 10.2 eV. After a time interval of the order of a microsecond, another photon collides with the same hydrogen atom with energy of 15 eV. What will be observed by the detector?

Answer 1

Hint We need to find the change in energy in the following states that are given in the question. This is a basic concept of modern physics.

Complete step by step answer:
We already know that the first photon which has energy 10.2eV will excite the hydrogen atom which is at ground state to the first excited state.
So, ${E_2} - {E_1} = 10.2eV$
Hence, during de-excitation a photon of 10.2 eV will be released. The second photon of energy 15 eV can ionise the atom. Hence the balance energy will be:
$15 - 13.6 = 1.4eV$
So, 1.4 eV is retained by the electron. Therefore, by the second photon an electron of energy 1.4 eV will be released.

Additional Information : Photon energy is the energy carried by a single photon. The amount of energy is directly proportional to the photon's electromagnetic frequency and thus, equivalently, is inversely proportional to the wavelength. The higher is the photon's frequency, the higher its energy. When an electron temporarily occupies an energy state greater than its ground state, it is in an excited state. An electron can become excited if it is given extra energy, such as if it absorbs a photon, or packet of light, or collides with a nearby atom or particle. The energy required to excite a hydrogen atom from n=1 to n=2 energy state 10.2 eV.

Note: It should be kept in mind that the energy is not the same in each state. When an electron is hit by a photon of light, it absorbs the quanta of energy the photon was carrying and moves to a higher energy state. One way of thinking about this higher energy state is to imagine that the electron is now moving faster.