Given below are two statements: one is labell | vedclass

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(D) The intensity of light $I$ is given by $I = \frac{n h 
u}{A}$,where $n$ is the number of photons per unit time,$h$ is Planck's constant,$
u$ is

Vedclass · Accepted Answer

$A$ is not correct but $R$ is correct.

Vedclass · Answer

(D) The intensity of light $I$ is given by $I = \frac{n h 
u}{A}$,where $n$ is the number of photons per unit time,$h$ is Planck's constant,$
u$ is the frequency,and $A$ is the area.From this,the number of photons per unit time is $n = \frac{IA}{h 
u}$.If the intensity $I$ is kept constant,increasing the frequency $
u$ results in a decrease in the number of photons $n$. Therefore,Assertion $A$ is incorrect.According to Einstein's photoelectric equation,the maximum kinetic energy $K_{\max}$ of emitted electrons is given by $K_{\max} = h 
u - \phi$,where $\phi$ is the work function.As the frequency $
u$ increases,the maximum kinetic energy $K_{\max}$ increases. Therefore,Reason $R$ is correct.

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