In the graph given below,if the slope is 4.12 | vedclass

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(B) According to Einstein's photoelectric equation,the kinetic energy of emitted electrons is given by $K_{max} = h
u - \phi$,where $h$ is Planck's c

Vedclass · Accepted Answer

$6.6 	imes 10^{-34} \, J-s$

Vedclass · Answer

(B) According to Einstein's photoelectric equation,the kinetic energy of emitted electrons is given by $K_{max} = h
u - \phi$,where $h$ is Planck's constant,$
u$ is the frequency of incident light,and $\phi$ is the work function.Since $K_{max} = eV_s$,where $V_s$ is the stopping potential and $e$ is the charge of an electron,we can write:$eV_s = h
u - \phi$$V_s = (\frac{h}{e})
u - \frac{\phi}{e}$Comparing this with the equation of a straight line $y = mx + c$,the slope $m$ is equal to $\frac{h}{e}$.Given that the slope is $4.12 	imes 10^{-15} \, V-s$ and the charge of an electron $e = 1.6 	imes 10^{-19} \, C$,we have:$\frac{h}{e} = 4.12 	imes 10^{-15}$$h = (4.12 	imes 10^{-15}) 	imes (1.6 	imes 10^{-19})$$h \approx 6.592 	imes 10^{-34} \, J-s \approx 6.6 	imes 10^{-34} \, J-s$.

In the graph given below,if the slope is $4.12 \times 10^{-15} \, V-s$,then the value of $'h'$ should be:

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