Home
Class 11
CHEMISTRY
Calculate the frequency of light emitted...

Calculate the frequency of light emitted when an electron drops from a higher to lower energy level of an atom and the difference between the two energy level is `35.64xx10^(-13)` erg. `[h=6.626xx10^(-27)erg*s]`

Text Solution

Verified by Experts

If `E_(1) and E_(2)` be the two energy levels, then the difference between their energies is given by, `E_(2)-E_(1)=hv or, Delta=hv`
[h=Planck's constannt and v=frequency of light emitted].
`therefore v=(DeltaE)/(h)=(35.64xx10^(-13)erg)/(6.626xx10^(-13)erg*s)=5.3804xx10^(14)s^(-1)`
Promotional Banner

Topper's Solved these Questions

  • STRUCTURE OF ATOM

    CHHAYA PUBLICATION|Exercise WARM UP EXERCISE|157 Videos

  • STRUCTURE OF ATOM

    CHHAYA PUBLICATION|Exercise QUESTION ANSWER ZONE FOR BOARD EXAMINATION (VERY SHORT ANSWER TYPE)|28 Videos

  • STATES OF MATTER : GASES AND LIQUIDS

    CHHAYA PUBLICATION|Exercise PRACTICE SET|10 Videos

Similar Questions

Explore conceptually related problems

When an electron of H-atom jumps from a higher to lower energy state, then-

Calculate the frequency nad wave length of the energy emitted when the electron jumps from 4th orbit to 1st orbit of a H-atom.

What is the wavelength of llight emitted when the electron in a hydrogen atom undergoes transition from an energy level with n=4 to an energy level with n=2 ?

How many photons are emitted in the transition of the electron from the fourth to the first energy level of H-atom?

Calculate the wave length in.angsfrom of the photon emitted when an electron returns to 2nd orbit from third orbit in the hydrogen atom. Given ionisation energy of hydrogen = 2.17 xx 10^(-11) erg /atom , h = 6.62 xx 10^(-27) erg sec.

Calculate the velocity of an electron having wavelength of 0.15 nm Mass of an electron is 9.109xx10^(-28) g (h=6.626 xx10^(-27) erg-s)

Calculate the velocity of an electron having wavelength of 0.15 nm Mass of an electron is 9.109xx10^(-28) g (h=6.626 xx10^(-27) erg-s)

Calculate the velocity of an electron having de Broglie wavelength of 200 Å. [m=9.11xx10^(-31)kg,h=6.626xx10^(-34)J*s] .

The radiation emitted when a hydrogen atom goes from a higher energy state to a lower energy state. The wavelength of one line in visible region of atomic spectrum of hydrogen is 6.63xx10^(-7) m. Energy difference between the two state is

Find the wavelength (in angstrom) of the photon emitted when an electron jumps from the second Bohr orbit to the first bohr orbit of hydrogen atom. Ionisation potential of hydrogen atom in its ground energy state =2.17xx10^(-11) erg *"atom"^(-1)