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yogesh gandhi Grade: 12
        

in the photo emissivecell with exicting wavelenght  X , THE MAX. kinetic energy of electron is K . if thr exicting wavelength is change to 3/4X the kinetic energy of the fastest emitted electron will be.....


1. equal to 3/4 k


2. 4/3K


3.  less than K.


4.  more than K.


 

7 years ago

Answers : (1)

Rajnish Mishra
23 Points
										


 


If the light frequency is greater than a threshold frequency, fo, then a photoelectron will be ejected with kinetic energy up to a maximum value, K, according to the linear formula:


 


Photoelectric Equation is given by: K = hc/λ – W


(where W is the work function of the metal surface, h and c constants)


 


If the exciting wavelength is changed by ¾ X then the equation for Kmax will be:


 


Kmax1 =   4hc/3X – W


 


We can see that the Kmax1 (that is new max kinetic energy of an electron) is greater than Kmax. Hence the kinetic energy of the fastest emitted electron will be more than K.


 


Answer – 4.


 



 



 


7 years ago
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