Home
Class 12
PHYSICS
Light of wavelength 2475 Å is incident o...

Light of wavelength 2475 Å is incident on barium . Photo electrons emitted describe a circle of radius 100 cm by a magnetic field of flux density `1/sqrt17xx10^(-5)` tesla. Find the work function of the barium . (Given `e/m=1.7xx10^11`)

Text Solution

Verified by Experts

Radius of circular path described by a charged particle in a magnetic field is given by
` t = sqrt((2mk)/(qB)) rArr k = (q^(2) B^(2) r^(2))/(2m) = (e/m) (eB^(2) r^(2))/(2) `
` = (1)/(2) xx 1.7 xx 10^(11) xx 1.6 xx 10^(-19) xx ((1)/(sqrt(17)) xx 10^(-5))^(2) xx (1)^(2)`
` 8 xx 10 ^(-20) J = 0.5 eV `
` E = phi + K_(max) `
` phi = E - K_(max)`
` = (12375)/(2475) eV - 0.5 eV = 4.5 eV `
Promotional Banner

Topper's Solved these Questions

  • ALTERNATING CURRENT

    AAKASH INSTITUTE ENGLISH|Exercise Try Yourself|15 Videos

  • ALTERNATING CURRENT

    AAKASH INSTITUTE ENGLISH|Exercise Assignment (Section - A) ( Objective Type Questions ( One option is correct))|40 Videos

  • ATOMS

    AAKASH INSTITUTE ENGLISH|Exercise ASSIGNMENT SECTION J (Aakash Challengers )|5 Videos

Similar Questions

Explore conceptually related problems

Light of wavelength 2475 Å is incident on barium. Photoelectrons emitted describe a circle of radius 100 cm by a magnetic field of flux density (1)/(sqrt(17)) xx 10^(-5) Tesla . Work function of the barium is (Given (e )/(m) = 1.7 xx 10^(11) )

Light of wavelength 4000Å is incident on barium. Photoelectrons emitted describe a circle of radius 50 cm by a magnetic field of flux density 5.26xx10^-6 tesla . What is the work function of barium in eV? Given h=6.6xx10^(-34)Js , e=1.6xx10^(-19)C, m_e=9.1xx10^(-31)kg.

When light of wavelangth 4000 Å is incident on a barium emitter , emitted photo electrons in a transverse magnetic field strength B = 5.2 xx 10^(-6) T moves in a circular path . Find the radius of the circle . Given work function of barium is 2.5 eV and mass of electron is 9.1 xx 10^(-31) kg . e = 1.6 xx 10^(-19) C , h = 6.6 xx 10^(34) Js .

When barium is irradiated by a light of lamda=4000A all the photoelectrons emitted are bent in a circle of radius 50 cm by a magnetic field of flux density 5.26xx10^(-6) T acting perpendicular to plane of emission of photoelectron. Then,

What is the strength of transverse magnetic field required to bend all the photoelectrons within a circle of a radius 50 cm when light of wavelength 3800 Å is incident on a barium emitter ? (Given that work function of barium is 2.5 eV, h=6.63xx10^(-34)js, e=1.6xx10^(19)C,m=9.1xx10^(-31)kg

Radiation falls on a target kept within a solenoid with 20 turns per cm, carrying a current 2.5 A. Electrons emitted move in a circle with a maximum radius of 1 cm. Find the wavelength of radiation, given that the work function of the target is 0.5 eV. q =1.6xx10^(-19)C , h = 6.625xx 10^(-34)J-s , mass of electron = 9.1 xx 10^(-31)kg

Photons of wavelength 6556 A^@ falls on a metal surface. If ejected electrons with maximum K.E. moves in magnetic field of 3 xx 10^(-4) T in circular orbit of radius 10^(-2) m, then work function of metal surface is

In a photocell bichromatic light of wavelength 2475 Å and 6000 Å are incident on cathode whose work function is 4.8 eV . If a uniform magnetic field of 3 xx 10^(-5) Tesla exists parallel to the plate , the radius of the path describe by the photoelectron will be (mass of electron = 9 xx 10^(-31) kg )

An electron of energy 1800 eV describes a circular path in magnetic field of flux density 0.4 T. The radius of path is (q = 1.6 xx 10^(-19) C, m_(e)=9.1 xx 10^(-31) kg)

Photo-electrons are produced from a metal surface using a radiation of wavelength 6561Å . These photo-electrons, when made to enter a uniform magnetic field of intensity 3 xx 10^(-4) T, move along different circular paths with a maximum radius of 10mm, then the work function of the metal is close to