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(B) According to the photoelectric effect,the number of photoelectrons emitted $(N)$ is directly proportional to the intensity (illumination $L$) of t

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$N \propto L$ and $E \propto \frac{1}{\lambda}$

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(B) According to the photoelectric effect,the number of photoelectrons emitted $(N)$ is directly proportional to the intensity (illumination $L$) of the incident light,provided the frequency is above the threshold frequency.According to Einstein's photoelectric equation,$E = \frac{hc}{\lambda} - W_0$,where $E$ is the maximum kinetic energy,$h$ is Planck's constant,$c$ is the speed of light,$\lambda$ is the wavelength,and $W_0$ is the work function of the metal.From this equation,it is clear that $E$ depends on the wavelength $\lambda$ such that $E \propto \frac{1}{\lambda}$ (assuming $W_0$ is constant).Therefore,$N \propto L$ and $E \propto \frac{1}{\lambda}$.

$A$ beam of light of wavelength $\lambda$ and with illumination $L$ falls on a clean surface of sodium. If $N$ photoelectrons are emitted each with kinetic energy $E$,then

Similar Questions

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