Electromagnetic radiation and spectra

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Question

The work function ( $ϕ$ ) of some metals is listed below. The number of metals which will show photoelectric effect when light of 300 nm wavelength falls on the metal is

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Solution

$The energy associated with incident photon = \frac{hc}{λ}$
$\begin{array}{l} \Rightarrow E = \frac{6 .6 \times 10^{- 34} \times 3 \times 10^{8}}{300 \times 10^{- 9}} J \\ E in eV = \frac{6 .6 \times 10^{- 34} \times 3 \times 10^{8}}{300 \times 10^{- 9} \times 1 .6 \times 10^{- 19}} = 4 .16 eV \end{array}$
So, number of metals showing photo-electric effects will be
(4), i.e., Li, Na, K, Mg

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Electromagnetic radiation and spectra

Remember concepts with our Masterclasses.

The work function (ϕ) of some metals is listed below. The number of metals which will show photoelectric effect when light of 300 nm wavelength falls on the metal is

The energy associated with incident photon =hcλ⇒E=6.6×10−34×3×108300×10−9JE in eV=6.6×10−34×3×108300×10−9×1.6×10−19=4.16eVSo, number of metals showing photo-electric effects will be (4), i.e., Li, Na, K, Mg

Ready to Test Your Skills?

free study material

The work function ( $ϕ$ ) of some metals is listed below. The number of metals which will show photoelectric effect when light of 300 nm wavelength falls on the metal is