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

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)`

A

4

B

6

C

2

D

1

Text Solution

Verified by Experts

The correct Answer is:
B
Promotional Banner

Similar Questions

Explore conceptually related problems

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.

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))

The ground state energy of hydrogen atom is -13.6eV . If the electron jumps to the ground state from the 3^("rd") excited state, the wavelength of the emitted photon 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 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 .

An excited He^(+) ion emits photon of wavelength lambda in returning to ground state from n^(th) orbit. If R is Rydberg's constant then :

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:

Find the wavenumber curresponding to the longest wavelength photon to remove electron from the second excited state of He^(o+) "ion" (R= 1.097 xx 10^(7)m^(-1))

The wavelength of radiation emitted when in He^(+) electron falls infinity to stationary state would be (R =1.098 xx 10 ^7 m^(-1))

A hydrogen atom in a state of binding energy 0.85 eV makes a transition to a state of excitation energy of 10.2 eV . Find the energy and wavelength of photon emitted.