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# The photo electric emission requires a threshold frequency $V_o$. For a certain metal $\lambda_1$ = 2200 $\overset{\circ}{A}$ and $\lambda_2$ = 1900 $\overset{\circ}{A}$ produce electrons with a maximum kinetic energy $KE_1$ and $KE_2$. If $KE_2 = 2KE_1$. Calculate $V_o$ and corresponding $\lambda_o$.

$\begin{array}{1 1} (a)\;V_o= 1.1483S^{-1} \lambda = 26126\;\overset{\circ}{A}\\ (b)\;V_o= 1.1483\times10^{15}S^{-1} \lambda = 2612.6\;\overset{\circ}{A}\\ (c)\;V_o= 2.438\times10^{12}S^{-1} \lambda = 2634\;\overset{\circ}{A}\\ (d)\;V_o= 1\times10^{15}S^{-1} \lambda = 3642\;\overset{\circ}{A}\end{array}$

Energy of photon = kinetic energy of photon electron + threshold frequency
$h\nu_1 = KE_1 + h\nu_o$ ---(1)
$h\nu_2 = KE_2 + h\nu_o$ -----(2)
Multiplying Eq(1) by (2) and subtracting Eq (2) from it
$2h\nu_1- h\nu_2 = h\nu_o$
$\nu_{\large\circ}$ = $[ \large\frac{2c}{\lambda_1} - \large\frac{c}{\lambda_2}]$ = $\large\frac{3\times10^8}{10^{-10}}$ [$\large\frac{2}{2200}-\large\frac{1}{1900}$]
also
$\lambda_{\large\circ}$ = $\large\frac{C}{\nu_o}$
= $\large\frac{3\times10^8}{1.1483\times10^{15}}$
$= 2.6126\times10^{-7}$ m
$=2612.6\overset{\circ}{A}$
edited Mar 19, 2014