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According to the photoelectric effect, the plot of kinetic energy of the emitted photo-electrons from a metal versus the frequency of the incident radiation gives a straight line whose slope
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is the same for all metals and independent of the intensity of radiation
For photoelectric effect, $\phi=\phi_0+(\mathrm{KE})_{\max }$
where, $\phi=$ energy of incident photon $=h v$
$\begin{aligned} & \begin{aligned} & \phi_0= \text { work function of material of metal on } \\ & \text { which photon is falling } \\ &(\mathrm{KE})_{\max }= \text { Maximum kinetic energy of emitted } \\ & \text { photoelectrons. }\end{aligned} \\ & \text { So, on putting the values } \\ & \qquad \begin{aligned} h v=\phi_0+(\mathrm{KE})_{\max } \\ \text { where, } v=\text { frequency of falling photon } \\ \text { and } h=\text { Planck's constant. }\end{aligned}\end{aligned}$
On comparing Eqs. (i) and (ii), we get $m=h=$ constant
So, the slope of $(\mathrm{KE})_{\max }$ versus $v$ graph is same for all metals and independent of the intensity of radiation.
where, $\phi=$ energy of incident photon $=h v$
$\begin{aligned} & \begin{aligned} & \phi_0= \text { work function of material of metal on } \\ & \text { which photon is falling } \\ &(\mathrm{KE})_{\max }= \text { Maximum kinetic energy of emitted } \\ & \text { photoelectrons. }\end{aligned} \\ & \text { So, on putting the values } \\ & \qquad \begin{aligned} h v=\phi_0+(\mathrm{KE})_{\max } \\ \text { where, } v=\text { frequency of falling photon } \\ \text { and } h=\text { Planck's constant. }\end{aligned}\end{aligned}$
On comparing Eqs. (i) and (ii), we get $m=h=$ constant
So, the slope of $(\mathrm{KE})_{\max }$ versus $v$ graph is same for all metals and independent of the intensity of radiation.
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