The work function of two photosensitive ...

The work function of two photosensitive surfaces A and B are 5 eV and 3 eV. Which surface should be selected to prepare a photo cell which is to be used for visible light ?

surface A

surface B

neither A nor B

Both A and B

Text Solution

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The correct Answer is:

`lambda=(hC)/(E)` , for `5 eV, lambda_(1)=2486 Å` (ultraviolet region )
For `3 eV, lambda_(2)=4143 Å` (Visible Region)
`:.` Surface B should be selected .

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