JEE MAIN - Physics (2020 - 5th September Evening Slot - No. 20)
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
2
Explanation
Given,
E1 = 4 eV
E2 = 2.5 eV
and $${{{{\left( {{V_1}} \right)}_{\max }}} \over {{{\left( {{V_2}} \right)}_{\max }}}} = 2$$
$${{{1 \over 2}m{{\left( {{{\left( {{V_1}} \right)}_{\max }}} \right)}^2}} \over {{1 \over 2}m{{\left( {{{\left( {{V_2}} \right)}_{\max }}} \right)}^2}}} = {{{E_1} - {\phi _0}} \over {{E_2} - {\phi _0}}}$$
$$ \Rightarrow $$ $${{{{\left( {{{\left( {{V_1}} \right)}_{\max }}} \right)}^2}} \over {{{\left( {{{\left( {{V_2}} \right)}_{\max }}} \right)}^2}}} = {{4 - {\phi _0}} \over {2.5 - {\phi _0}}}$$
$$ \Rightarrow $$ $${\left( 2 \right)^2} = {{4 - {\phi _0}} \over {2.5 - {\phi _0}}}$$
$$ \Rightarrow $$ 10 - 4$$\phi $$0 = 4 - $$\phi $$0 $$ \Rightarrow $$ 3$$\phi $$0 = 6
$$ \Rightarrow $$ $$\phi $$0 = 2
$$ \therefore $$ Work function($$\phi $$) of the metal = 2 eV
E1 = 4 eV
E2 = 2.5 eV
and $${{{{\left( {{V_1}} \right)}_{\max }}} \over {{{\left( {{V_2}} \right)}_{\max }}}} = 2$$
$${{{1 \over 2}m{{\left( {{{\left( {{V_1}} \right)}_{\max }}} \right)}^2}} \over {{1 \over 2}m{{\left( {{{\left( {{V_2}} \right)}_{\max }}} \right)}^2}}} = {{{E_1} - {\phi _0}} \over {{E_2} - {\phi _0}}}$$
$$ \Rightarrow $$ $${{{{\left( {{{\left( {{V_1}} \right)}_{\max }}} \right)}^2}} \over {{{\left( {{{\left( {{V_2}} \right)}_{\max }}} \right)}^2}}} = {{4 - {\phi _0}} \over {2.5 - {\phi _0}}}$$
$$ \Rightarrow $$ $${\left( 2 \right)^2} = {{4 - {\phi _0}} \over {2.5 - {\phi _0}}}$$
$$ \Rightarrow $$ 10 - 4$$\phi $$0 = 4 - $$\phi $$0 $$ \Rightarrow $$ 3$$\phi $$0 = 6
$$ \Rightarrow $$ $$\phi $$0 = 2
$$ \therefore $$ Work function($$\phi $$) of the metal = 2 eV
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