JEE MAIN - Physics (2002 - No. 23)
Sodium and copper have work functions $$2.3$$ $$eV$$ and $$4.5$$ $$eV$$ respectively. Then the ratio of the wavelengths is nearest to
$$1:2$$
$$4:1$$
$$2:1$$
$$1:4$$
Explanation
We know that work function is the energy required and energy $$E = h\upsilon $$
$$\therefore$$ $${{{E_{NA}}} \over {{E_{Cu}}}} = {{h{\upsilon _{Na}}} \over {h{\upsilon _{cu}}}} = {{{\lambda _{cu}}} \over {{\lambda _{Na}}}}$$
$$\left[ {\,\,} \right.$$ as $${\,\,\,\upsilon \propto {1 \over \lambda }}$$ for light $$\left. {\,\,} \right]$$
$$\therefore$$ $${{{\lambda _{Na}}} \over {{\lambda _{Cu}}}} = {{{E_{Cu}}} \over {{E_{Na}}}} = {{4.5} \over {2.3}} \approx {2 \over 1}$$
$$\therefore$$ $${{{E_{NA}}} \over {{E_{Cu}}}} = {{h{\upsilon _{Na}}} \over {h{\upsilon _{cu}}}} = {{{\lambda _{cu}}} \over {{\lambda _{Na}}}}$$
$$\left[ {\,\,} \right.$$ as $${\,\,\,\upsilon \propto {1 \over \lambda }}$$ for light $$\left. {\,\,} \right]$$
$$\therefore$$ $${{{\lambda _{Na}}} \over {{\lambda _{Cu}}}} = {{{E_{Cu}}} \over {{E_{Na}}}} = {{4.5} \over {2.3}} \approx {2 \over 1}$$
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