The surface of a metal is illuminated alternately with photons of energies E1 = 4 eV and E2 = 2.5 eV respectively. The ratio of maximum speeds of the photoelectrons emitted in the two cases is 2. The work function of the metal in (eV) is _____.
Answer (integer)
2
Solution
Given,
<br><br>E<sub>1</sub> = 4 eV
<br><br>E<sub>2</sub> = 2.5 eV
<br><br>and $${{{{\left( {{V_1}} \right)}_{\max }}} \over {{{\left( {{V_2}} \right)}_{\max }}}} = 2$$
<br><br>$${{{1 \over 2}m{{\left( {{{\left( {{V_1}} \right)}_{\max }}} \right)}^2}} \over {{1 \over 2}m{{\left( {{{\left( {{V_2}} \right)}_{\max }}} \right)}^2}}} = {{{E_1} - {\phi _0}} \over {{E_2} - {\phi _0}}}$$
<br><br>$\Rightarrow$ $${{{{\left( {{{\left( {{V_1}} \right)}_{\max }}} \right)}^2}} \over {{{\left( {{{\left( {{V_2}} \right)}_{\max }}} \right)}^2}}} = {{4 - {\phi _0}} \over {2.5 - {\phi _0}}}$$
<br><br>$\Rightarrow$ ${\left( 2 \right)^2} = {{4 - {\phi _0}} \over {2.5 - {\phi _0}}}$
<br><br>$\Rightarrow$ 10 - 4$\phi$<sub>0</sub> = 4 - $\phi$<sub>0</sub>
$\Rightarrow$ 3$\phi$<sub>0</sub> = 6
<br><br>$\Rightarrow$ $\phi$<sub>0</sub> = 2
<br><br>$\therefore$ Work function($\phi$) of the metal = 2 eV
About this question
Subject: Physics · Chapter: Dual Nature of Matter and Radiation · Topic: Photoelectric Effect
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