If $e/m$ of electron is $1.76 imes 10^{11}$ C ${kg}^{- 1}$ and stopping potential is 0.71 V, then the maximum velocity of the photoelectron is

Question

If $e/m$ of electron is $1.76 	imes 10^{11}$ C ${kg}^{- 1}$ and stopping potential is 0.71 V, then the maximum velocity of the photoelectron is

Dr. Abubacker · Accepted Answer

(c) $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{1}{2}mv^{2} = eV$
$v\  = \sqrt{\frac{2eV}{m}} = \sqrt{2V 	imes \frac{e}{m}}$
$v = \ \sqrt{2 	imes 0.91 	imes 1.76 	imes 10^{11}}$
$\ \  = 5 	imes 10^{5}m/s$
= 500Km/s

Q. If $e/m$ of electron is $1.76 \times 10^{11}$ C ${kg}^{- 1}$ and stopping potential is 0.71 V, then the maximum velocity of the photoelectron is

Explanation

Reviewed by Dr. Abubacker

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