Subjective Type

The work function of a metal is $$2\cdot 5\times { 10 }^{ -9 }J$$. (a) Find the threshold frequency for photoelcectric emission. (b) If the metal is exposed to a light beam of frequency $$6\cdot 0\times { 10 }^{ 14 }Hz$$, what will be the stopping potential ?

Solution

$$W_0=2.5\times10^{-19}J$$

a) We know $$W_0=h\nu_0$$

$${ \nu}_{ 0 }=\dfrac { { W }_{ 0 } }{ h } $$

$$=\dfrac { 2.5\times { 10 }^{ -19 } }{ 6.63\times { 10 }^{ -34 } } $$

$$=3.77\times { 10 }^{ 14 }Hz=3.8\times { 10 }^{ 14 }Hz$$

b) The stopping potential is obtained as:
$$eV_0=h\nu-W_0$$

or, $$V_0=\dfrac { h\nu-{ W }_{ 0 } }{ e } $$

$$=\dfrac { 6.63\times { 10 }^{ -34 }\times 6\times { 10 }^{ 14 }-2.5\times { 10 }^{ -19 } }{ 1.6\times { 10 }^{ -19 } } =0.91V$$

SIMILAR QUESTIONS

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