Home
Class 12
PHYSICS
Find the quantum number n corresponding ...

Find the quantum number n corresponding to nth excited state of `He^(++)` ion if on transition to the ground state the ion emits two photons in succession with wavelength 108.5 nm and 30.4 nm. The ionization energy of the hydrogen atom is 13.6 eV.

Text Solution

Verified by Experts

The correct Answer is:
5
Promotional Banner

Similar Questions

Explore conceptually related problems

Find the quantum number n corresponding to the excited state of He^(+) ion if on transition to the ground state that ion emits two photons in succession with wavelengths state that ion emits two photons in succession with wavelengths 1026.7 and 304Å. (R = 1.096 xx 10^(7)m^(_1)

Find the quantum of the excited state of electrons in He^(+) ion which on transition to first excited state emit photons of wavelengths 108.5 nm . (R_(H)=1.09678xx10^(7) m^(-1))

A 100 eV electron collides with a stationary helium ion (He^(+)) in its ground state and excites to a higher level. After the collision , He^(+) ion emits two photons in succession with wavelength 1085 Å and 304 Å . Find the principal quantum number of the excite in its ground state and. Also calculate energy of the electron after the collision. Given h = 6.63 xx 10^(-34) J s .

Consider a hydrogen-like ionized atom with atomic number with a single electron. In the emission spectrum of this atom, the photon emitted in the 2 to 1 transition has energy 74.8 higher than the photon emitted in the 3 to 2 transition. The ionization energy of the hydrogen atom is 13.6. The value of Z is __________.

He^(+) is in n^(th) state. It emits two successive photons of wavelength 103.7nm and 30.7nm , to come to ground state the value of n is:

A hydrogen like atom (atomic number Z) is in a higher excited state of quantum number n. The excited atom can make a transition ot the first excited state by successively emitting two photons of energy 10.2 eV and 17.0 eV, respectively. Alternatively, the atom from the same excited state can make a transition to the second excited state by successively emitting two photons of energies 4.25 eV and 5.95 eV, respectivley Determine the values of n and Z. (lonization energy of H-atom = 13.6 eV)

How many different photons can be emitted by hydrogen atoms that undergo transitions to the ground state from the n=5 states?

In atension of state n from a state of excitation energy given is 10.19 eV , hydrogen atom emits a photon with wavelength 4890 Å . Determine the binding energy of the initial state.

A hydrogen like atom (atomic number z ) is in a higher excited state of quantum number n . This excited atom can make a transition to the first excited state by successively emitting two photons of energies 10.2 eV and 17.0 eV respectively. Alternatively the atom from the same excited state can make a transition to the second excited state by successively emitting 2 photons of energy 4.25 eV and 5.95 eV respectively. Determine the value of (n+z)

A hydrogen like species (atomic number Z) is present in a higher excited state of quantum number n. This excited atom can make a transitionn to the first excited state by successive emission of two photons of energies 10.20 eV and 17.0 eV respectively. Altetnatively, the atom from the same excited state can make a transition to the second excited state by successive of two photons of energy 4.25 eV and 5.95 eVv respectively. Determine the value of Z.