Home
Class 12
PHYSICS
A photelectric meterial having work-func...

A photelectric meterial having work-function `phi_(0)` is illuminated with light of wavelegth `lamda(lamdalt(hc)/(4_(0))).` The fastes photoelectron has a de Broglie wevelength `lamda_(d).` A change in wavelength of the incident light by `Deltalamda` results in a change `Deltalamda_(d)` in `lamda_(d).` Then the ration `Deltalamda_(d)//Deltalamda` is proportional to

A

`(lambda_(d)^(3))/(lambda^(2))`

B

`(lambda_(d)^(3))/lambda`

C

`(lambda_(d)^(2))/(lambda^(2))`

D

`(lambda_(d))/lambda`

Text Solution

Verified by Experts

The correct Answer is:
A
`(hc)/lambda =W_(0)+KE_(max)`
`KE =(P^(2))/(2m_(e))=(h^(2))/(2m_(e)lambda_(d)^(2))`
`(hc)/lambda =phi_(0)+(h^(2))/(2m_(e)lambda_(d)^(2))`
`-(hc)/(lambda^(2))dlambda=0+(h^(2))/(2m_(e))((-2))/(lambda_(d)^(3))dlambda_(d)`
`(dlambda_(d))/(dlambda)=(m_(e)lambda_(d)^(3))/(h^(2)xxlambda^(2))hc`
`(dlambda_(d))/(dlambda)prop (lambda_(d)^(3))/(lambda^(2))` .
Promotional Banner

Similar Questions

Explore conceptually related problems

A photoelectric material having work-function phi_(0) is illuminated with light of wavelength lamda(lamdalt(hc)/(4_(0))). The fastest photoelectron has a de Broglie wavelength lamda_(d). A change in wavelength of the incident light by Deltalamda results in a change Deltalamda_(d) in lamda_(d). Then the ration Deltalamda_(d)//Deltalamda is proportional to

Electrons are emitted from the cathode of a photocell of negligible work function, when photons of wavelength lamda are incident on it. Derive the expression for the de Broglie wavelength of the electrons emitted in terms of the wavelength of the incident light

Four elements A, B, C, D have work function 2, 2.4, 2.8, 3.2 eV. Light of wavelength 4000 Å is incident on them. The elements which emit photo electrons are

In a photomissive cell with executing wavelength lamda//2 , work function phi , the fastest moving electron has acquired speed as v. If the executing wavelength is changed to (5lamda)/4 , the speed of the fastest emitted electron will be

If photoelectrons emitted by a metl on irradiating it by a light of wave length lamda cannot move away farther than a distance 'd' in presence of a retarding electric field E, find threshold wavelength of the metal.

From hydrogen gas discharge tube, light is emitted. Emitted light is used to emit electron from sodium metal of work function phi= 1.82 eV If the fastest photoelectron has KE of 0.73 eV, the wavelength of incident radiation on the sodium metal is :

Photoelectrons ar ejected from a surface when light of wavelenght lamda_1=550 nm is incident on it. The stopping potential for such electrons is lamda_(S1)=0.19V . Suppose that radiation of wavelength lamda_2=190nm is incident of the surface. Q. Calulate the work function of the surface.

In a Young's double slit experiment, if the incident light consists of two wavelengths lamda_(1)andlamda_(2) , the slit separation is d, and the distance between the slit and the screen is D, the maxima due to the two wavelengths will coincide at a distance from the central maxima, given by :