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The photoelectric cut-off voltage in a certain experiment is 1.5 V. What is the maximum kinetic energy of photoelectrons emitted ?

$\begin {array} {1 1} (a)\;2.4 \times 10^{-19}J & \quad (b)\;24 \times 10^{-19}J \\ (c)\;240 \times 10^{-19}J & \quad (d)\;2.4 \times 10^{-18}J \end {array}$

 

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1 Answer

Photoelectric cut-off voltage, $V_{\circ}=1.5V$
$K_e =eV_{\circ}\: \; \: $ where,$e$ = charge on an electron $ = 1.6 \times 10^{-19}C$
So, $ K_e = 1.6 \times 10^{-19} \times 1.5 = 2.4 \times 10^{-19}J$

Ans : (a)

answered Dec 23, 2013 by thanvigandhi_1
edited Mar 13, 2014 by thanvigandhi_1
 

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