JEE MAIN - Physics (2016 (Offline) - No. 3)
Radiation of wavelength $$\lambda ,$$ is incident on a photocell. The fastest emitted electron has speed $$v.$$ If the wavelength is changed to $${{3\lambda } \over 4},$$ the speed of the fastest emitted electron will be:
$$ = v{\left( {{4 \over 3}} \right)^{{1 \over 2}}}$$
$$ = v{\left( {{3 \over 4}} \right)^{{1 \over 2}}}$$
$$ > v{\left( {{4 \over 3}} \right)^{{1 \over 2}}}$$
$$ < v{\left( {{4 \over 3}} \right)^{{1 \over 2}}}$$
Explanation
$$h{v_0}^2 - h{v_0} = {1 \over 2}m{v^2}$$
$$\therefore$$ $${4 \over 3}h{v_0} - h{v_0} = {1 \over 2}mv{'^2}$$
$$\therefore$$ $${{v{'^2}} \over {{v^2}}} = {{{4 \over 3}v - {v_0}} \over {v - {v_0}}}$$
$$\therefore$$ $$v' = v\sqrt {{{{4 \over 3}v - {v_0}} \over {v - {v_0}}}} $$
$$\therefore$$ $$v' > v\sqrt {{4 \over 3}} $$
$$\therefore$$ $${4 \over 3}h{v_0} - h{v_0} = {1 \over 2}mv{'^2}$$
$$\therefore$$ $${{v{'^2}} \over {{v^2}}} = {{{4 \over 3}v - {v_0}} \over {v - {v_0}}}$$
$$\therefore$$ $$v' = v\sqrt {{{{4 \over 3}v - {v_0}} \over {v - {v_0}}}} $$
$$\therefore$$ $$v' > v\sqrt {{4 \over 3}} $$
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