When radiation is incident on a photoelectron | vedclass

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(C) The maximum kinetic energy of the ejected photoelectrons is related to the stopping potential $V_0$ by the equation: $K_{\max} = \frac{1}{2}mv_{\m

Vedclass · Accepted Answer

$1.8 	imes 10^6 \ m \ s^{-1}$

Vedclass · Answer

(C) The maximum kinetic energy of the ejected photoelectrons is related to the stopping potential $V_0$ by the equation: $K_{\max} = \frac{1}{2}mv_{\max}^2 = eV_0$. Rearranging this equation to solve for the maximum velocity $v_{\max}$,we get: $v_{\max} = \sqrt{2(\frac{e}{m})V_0}$. Given values: $V_0 = 9 \ V$ and $\frac{e}{m} = 1.8 	imes 10^{11} \ C \ kg^{-1}$. Substituting these values into the formula: $v_{\max} = \sqrt{2 	imes (1.8 	imes 10^{11}) 	imes 9}$. $v_{\max} = \sqrt{32.4 	imes 10^{11}} = \sqrt{3.24 	imes 10^{12}} = 1.8 	imes 10^6 \ m \ s^{-1}$.

When radiation is incident on a photoelectron emitter,the stopping potential is found to be $9 \ V$. If $e/m$ for the electron is $1.8 \times 10^{11} \ C \ kg^{-1}$,the maximum velocity of the ejected electrons is:

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