If the maximum velocity with which an electro | vedclass

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

Vedclass · Accepted Answer

$45$

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

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

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