Question

The work function (Ф) 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

Answer 1

for energy in eV,

$e = \frac{1240}{\lambda}ev \\ e = \frac{1240}{300} = 4.13ev$

so for excitation of electron ,

$e \geqslant \phi$

so the metals which will show excitation are Cu,Ag, Fe, Pt,W

Answer is 5

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Answer 2

Answer:

E=HC/λ

 =6.6×10  POWER −34  × 3×10 POWER 8/300×10  POWER −9 J

 E(in eV)= 6.6×10  −34 ×3×10  POWER 8  /300×10  POWER  −9  ×1.6×10 POWER   −19  =4.125 eV

For photoelectric effect to occur, the energy of a photon must be greater than the work function of the metal.

So, the number of metals showing photo-electric effects will be (4) i.e., Li, Na, K, Mg.