Single Choice

The threshold frequency for a photo-sensitine metal is $$3.3\times { 10 }^{ 14 }Hz$$. If light of frequency $$8.2\times { 10 }^{ 14 }Hz$$ is incident on this metal, the cut-off voltage for the photo-electric emission is nearly

A$$2 V$$
Correct Answer
B$$3 V$$
C$$5 V$$
D$$1 V$$

Solution

Here, $${ V }_{ 0 }=\dfrac { E-V }{ e } =\dfrac { h\left( v-{ v }_{ 0 } \right) }{ e }$$
$$ =\dfrac { 6.62\times { 10 }^{ -34 }\left( 8.2\times { 10 }^{ 14 }-3.3\times { 10 }^{ 14 } \right) }{ 1.6\times { 10 }^{ -19 } }$$
$$ =\dfrac { 6.62\times { 10 }^{ -34 } }{ 1.6 } \times 4.9\times { 10 }^{ 33 }$$
$$=\dfrac { 6.62\times 4.9\times { 10 }^{ -1 } }{ 1.6 }$$
$$ { V }_{ 0 }=2$$ volt

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