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  • Figure shows the stopping potential for the photo electron versus

Figure shows the stopping potential (V_0 ) for the photo electron versus \left (\frac{1}{\lambda } \right ) graph, for two metals A and B, \lambda being the wavelength of incident light.

(a) How is the value of Planck’s constant determined from the graph ?

(b) If the distance between the light source and the surface of metal A is increased, how will the stopping potential for the electrons emitted from it be effected ? Justify your answer.

 

 
 
 
 
 

Answers (1)

a) 

\\h\nu=\phi_0+eV_s\\eV_s=h\nu-\phi_0\\V_s=\frac{h\nu}{e}-\frac{\phi_0}{e}\\\\V_s=(\frac{hc}{e})\frac{1}{\lambda}+(\frac{-\phi_0}{e})

Comparing with y=mx+c

m=\frac{hc}{e}

h=\frac{me}{c}

(b) Stopping potential will remain the same
Variation of the distance of the light source from the metal surface will alter the
intensity while the stopping potential, however, depends only on the frequency
and not on the intensity of the incident light.

Posted by

Safeer PP

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