Home
Class 12
CHEMISTRY
A sample of hydrogen gas has same atom i...

A sample of hydrogen gas has same atom in out excited state and same atom in other excited state it emits three difference photon.When the sample was irradiated with radiation of energy `2.85 eV` ,it emits `10` different photon all having energy in or less than `13.6 eV` ltbrtgt a. Find the principal quantum number of initially excited electrons

Text Solution

Verified by Experts

Initially sample emit three different photons. It means for one of the excited states, n = 3.
For other excited state, n = 1 or n = 2
After irradiation it emits 10 different photons, it means for this excited state, n = 5
Energy of `5^(th)` energy level ` = ( - 13.6)/((5)^(2)) = - 0.544 eV`
Energy of `3^(rd)` energy level ` = ( - 13.6)/((3)^(2)) = - 1.51 eV`
Energy of `2^(nd)` energy level ` = ( 13.6)/((2)^(2)) = - 3.4 e V`
So , after absorption of 2.85eV energy, only electron from second energy level can jump to `5^(th)` energy level. Principal quantum number of initially excited state are 2 and 3 respectively.
Maximum energy of emitted photon `= 13.6 - 15.1 = 12.09eV`
Minimum energy of emitted photon = 3.4 - 1.51 = 1.89 eV
Promotional Banner

Similar Questions

Explore conceptually related problems

An electron in nth excited state in a hydrogen atom comes down to first excited state by emitting ten different wavelength. Find value of n (an integer).

The energy of an atom or ion in the first excited state is -13.6 eV. It may be

A gas of identical hydrogen-like atoms has some atoms in the lowest in lower (ground) energy level A and some atoms in a partical upper (excited) energy level B and there are no atoms in any other energy level.The atoms of the gas make transition to higher energy level by absorbing monochromatic light of photon energy 2.7 e V . Subsequenty , the atom emit radiation of only six different photon energies. Some of the emitted photons have energy 2.7 e V some have energy more , and some have less than 2.7 e V . a Find the principal quantum number of the intially excited level B b Find the ionization energy for the gas atoms. c Find the maximum and the minimum energies of the emitted photons.

An excited state of H atom emits a photon of wavelength lamda and returns in the ground state. The principal quantum number of excited state is given by:

In a hydrogen atom, if energy of an electron in ground state is - 13.6 eV , then that in the 2^(nd) excited state is :

Photons of equal energy were incident on two different gas samples. One sample containing H-atoms in the ground state and the other sample containing H-atoms in some excited state with a principle quantum number 'n'. The photonic beams totally ionise the H-atoms. If the difference in the kinetic energy of the ejected electrons in the two different cases is 12.75 eV . Then find the principal quantum number 'n' of the excited state.

In hydrogen atom, energy of first excited state is - 3.4 eV . Then, KE of the same orbit of hydrogen atom is.

Energy of H-atom in the ground state is -13.6 eV, hence energy in the second excited state is

A gaseous excited hydrogen-like species with nuclear charge Z can emit radiations of six different photon energies. The principal quantum number of the excited state is :

The electron of H-atom in the ground state is excited to a higher energy level by monuchromatic light of energy 13.22 eV How many different photon are emitted when it return to the ground state?