When monochromatic light falls on a photo-sen | vedclass

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(A) The maximum kinetic energy $(K.E.)_{\max}$ of the emitted electron is given by the formula:$(K.E.)_{\max} = \frac{1}{2} m v_{\max}^2$Given $m = 9

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$7.2$

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(A) The maximum kinetic energy $(K.E.)_{\max}$ of the emitted electron is given by the formula:$(K.E.)_{\max} = \frac{1}{2} m v_{\max}^2$Given $m = 9 	imes 10^{-31} \ kg$ and $v_{\max} = 1.6 	imes 10^6 \ m/s$,we have:$(K.E.)_{\max} = \frac{1}{2} 	imes (9 	imes 10^{-31}) 	imes (1.6 	imes 10^6)^2$$(K.E.)_{\max} = 0.5 	imes 9 	imes 10^{-31} 	imes 2.56 	imes 10^{12}$$(K.E.)_{\max} = 11.52 	imes 10^{-19} \ J$The stopping potential $V_s$ is related to the maximum kinetic energy by the equation:$e V_s = (K.E.)_{\max}$$V_s = \frac{(K.E.)_{\max}}{e} = \frac{11.52 	imes 10^{-19} \ J}{1.6 	imes 10^{-19} \ C}$$V_s = 7.2 \ V$Therefore,the correct option is $A$.

When monochromatic light falls on a photo-sensitive metal,an electron is emitted with maximum velocity $1.6 \times 10^6 \ m/s$. Find the stopping potential.
[charge of electron $= 1.6 \times 10^{-19} \ C$,mass of electron $= 9 \times 10^{-31} \ kg$] (in $V$)

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When monochromatic light falls on a photo-sensitive metal,an electron is emitted with maximum velocity $1.6 \times 10^6 \ m/s$. Find the stopping potential.[charge of electron $= 1.6 \times 10^{-19} \ C$,mass of electron $= 9 \times 10^{-31} \ kg$] (in $V$)