Home
Class 11
CHEMISTRY
Light of wavelength 5000 Å fall on a me...

Light of wavelength `5000 Å` fall on a metal surface of work function `1.9` eV. Find
a. The energy of photon
b. The kinetic energy of photoelectrons

Text Solution

Verified by Experts

Wavelength of light `(lambda) = 5000 Ã… = 5000 xx 10^(-10)m =5 xx 10^(-7)m`.
Work function `(v^(0))=1.9eV=1.9 xx 1.6 xx 10^(-19) J`.
Step I. Energy of photons `E=hv=(hc)/(lambda)=((6.626 xx 10^(-34)Js)xx (3 xx 10^(8) ms^(-1)))/((5xx 10^(-7)m))=3.97 xx 10^(-19)J`.
Step II . Kinetic energy of photoelectrons ltbr. `KE(1//2 mv^(2))=hv-hv^(0)=(3.97 xx 10^(-19)J)-(1.9 xx 1.6 xx 10^(-19)J) `
`=(3.97 xx 10^(19)J)-(3.04 xx 10^(-19)J)=9.3 xx 10^(-20)J`.
Promotional Banner

Topper's Solved these Questions

  • STRUCTURE OF ATOM

    DINESH PUBLICATION|Exercise Board Examinations|67 Videos

  • STRUCTURE OF ATOM

    DINESH PUBLICATION|Exercise Short Answer type Questions|24 Videos

  • STATES OF MATTER : GASES AND LIQUIDS

    DINESH PUBLICATION|Exercise Statement Type Question|5 Videos

  • SURFACE CHEMISTRY

    DINESH PUBLICATION|Exercise Ultimate Prepatarory Package|16 Videos

Similar Questions

Explore conceptually related problems

Light of wavelenght 5000 Å fall on a metal surface of work function 1.9 eV Find a. The energy of photon b. The kinetic energy of photoelectrons

Light of wavelength 5000Å falls on a metal surface of work function 1.9eV. Find (i) the energy of photons in eV (ii) the kinetic energy of photoelectrons and (iii) the stopping potential. Use h=6.63xx10^(-34)Js , c=3xx10^8ms^-1 , e=1.6xx10^-9C.

A photon of wavelength 5000 lambda strikes a metal surface with work function 2.20 eV calculate aTHe energy of the photon in eV b. The kinetic energy of the emitted photo electron c. The velocity of the photoelectron

Light of wavelength 2000Å is incident on a metal surface of work function 3.0 eV. Find the minimum and maximum kinetic energy of the photoelectrons.

Light of wavelength 2000 Å falls on a metallic surface whose work function is 4.21 eV. Calculate the stopping potential

Light of wavelength 5000 Å falls on a sensitive plate with photoelectric work function of 1.9 eV . The kinetic energy of the photoelectron emitted will be

If light of wavelength 6200Å falls on a photosensitive surface of work function 2 eV, the kinetic energy of the most energetic photoelectron will be

A light of wavelength 5000 Å falls on a photosensitive plate with photoelectric work function 1.90 eV. Kinetic energy of the emitted photoelectrons will be (h = 6.63 xx 10^(-34)J.s)

Photons of energy 1.5 eV and 2.5 eV are incident on a metal surface of work function 0.5 eV. What is the ratio of the maximum kinetic energy of the photoelectrons ?