JEE MAIN - Physics (2019 - 12th January Evening Slot - No. 11)
When a certain photosensitive surface is illuminated with monochromatic light of frequency v, the stopping
potential for the current is –V0/2. When the surface is illuminated by monochromatic light of frequency v/2, the stopping potential is – V0. The threshold frequency for photoelectric emission is :
2$$v$$
$${4 \over 3}v$$
$${{3v} \over 2}$$
$${{5v} \over 3}$$
Explanation
Einstein’s photoelectric equation in the two cases
is given by
$${{e{V_0}} \over {\Delta E}} = h\upsilon - h{\upsilon _0}$$ ......(i)
and $$e{V_0} = {{h\upsilon } \over 2} - h{\upsilon _0}$$ .....(ii)
From eqn. (i) and (ii),
$${1 \over 2} = {{h\upsilon - h{\upsilon _0}} \over {{{h\upsilon } \over 2} - h{\upsilon _0}}}$$
$$ \Rightarrow $$ $${\upsilon _0} = {3 \over 2}\upsilon $$
$${{e{V_0}} \over {\Delta E}} = h\upsilon - h{\upsilon _0}$$ ......(i)
and $$e{V_0} = {{h\upsilon } \over 2} - h{\upsilon _0}$$ .....(ii)
From eqn. (i) and (ii),
$${1 \over 2} = {{h\upsilon - h{\upsilon _0}} \over {{{h\upsilon } \over 2} - h{\upsilon _0}}}$$
$$ \Rightarrow $$ $${\upsilon _0} = {3 \over 2}\upsilon $$
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