If the maximum velocity with which an electro | vedclass

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(B) Given,maximum velocity $v_{\max} = 4 	imes 10^8 \,cm/sec = 4 	imes 10^6 \,m/sec$.Mass of electron $m = 9 	imes 10^{-31} \,kg$.Charge of electro

Vedclass · Accepted Answer

$45$

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(B) Given,maximum velocity $v_{\max} = 4 	imes 10^8 \,cm/sec = 4 	imes 10^6 \,m/sec$.Mass of electron $m = 9 	imes 10^{-31} \,kg$.Charge of electron $e = 1.6 	imes 10^{-19} \,C$.The maximum kinetic energy is given by $K_{\max} = \frac{1}{2} m v_{\max}^2$.$K_{\max} = \frac{1}{2} 	imes (9 	imes 10^{-31} \,kg) 	imes (4 	imes 10^6 \,m/sec)^2$.$K_{\max} = 0.5 	imes 9 	imes 10^{-31} 	imes 16 	imes 10^{12} = 72 	imes 10^{-19} \,J$.To convert this into electron-volts $(eV)$,divide by $e$:$K_{\max} = \frac{72 	imes 10^{-19} \,J}{1.6 	imes 10^{-19} \,C} = 45 \,eV$.The stopping potential $V_0$ is defined as $K_{\max} = e V_0$,so $V_0 = \frac{K_{\max}}{e} = 45 \,V$.

If the maximum velocity with which an electron can be emitted from a photocell is $4 \times 10^8 \,cm/sec$,the stopping potential is ............ $V$ (mass of electron $= 9 \times 10^{-31} \,kg$).

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