JEE MAIN - Chemistry (2019 - 11th January Evening Slot - No. 18)
The de Broglie wavelength ($$\lambda $$) associated with a photoelectron varies with the frequency (v) of the incident radiation as, [v0 is threshold frequency] :
$$\lambda \,\infty \,{1 \over {{{\left( {v - {v_0}} \right)}^{{3 \over 2}}}}}$$
$$\lambda \,\infty \,{1 \over {{{\left( {v - {v_0}} \right)}^{{1 \over 4}}}}}$$
$$\lambda \,\infty \,{1 \over {\left( {v - {v_0}} \right)}}$$
$$\lambda \,\infty \,{1 \over {{{\left( {v - {v_0}} \right)}^{{1 \over 2}}}}}$$
Explanation
By photoelectric effect
KE = h$$\gamma $$ - h$$\gamma $$o ....(1)
de broglie wavelength,
$$\lambda $$ = $${h \over {mv}}$$ = $${h \over {\sqrt {2m \times K.E} }}$$ ...(2)
Using equation (1) and (2), we get
$$\lambda $$ = $${h \over {\sqrt {2m \times \left( {h\nu - h{\nu _0}} \right)} }}$$
$$ \therefore $$ $$\lambda \,\propto \,{1 \over {{{\left( {v - {v_0}} \right)}^{{1 \over 2}}}}}$$
KE = h$$\gamma $$ - h$$\gamma $$o ....(1)
de broglie wavelength,
$$\lambda $$ = $${h \over {mv}}$$ = $${h \over {\sqrt {2m \times K.E} }}$$ ...(2)
Using equation (1) and (2), we get
$$\lambda $$ = $${h \over {\sqrt {2m \times \left( {h\nu - h{\nu _0}} \right)} }}$$
$$ \therefore $$ $$\lambda \,\propto \,{1 \over {{{\left( {v - {v_0}} \right)}^{{1 \over 2}}}}}$$
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