The work function of a photometal is 6.63 eV. The threshold wavelength is

  1. 3920 Å
  2. 1866 Å
  3. 186.6 Å
  4. 18666 Å

Answer (Detailed Solution Below)

Option 2 : 1866 Å
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Detailed Solution

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CONCEPT:

Work function (Φo):

  • The minimum energy of incident radiation, required to eject the electrons from the metallic surface is defined as the work function of that surface.

\(\Rightarrow {{\rm{\Phi }}_0} = h{\nu _0} = \frac{{hc}}{{{\lambda _0}}}\)

Where h = Plank's constant, c = speed of light, ν0 = threshold frequency and λ0 = threshold wavelength.              

EXPLANATION:

Given - Work function (Φo) = 6.63 eV

  • The threshold wavelength is

\(\Rightarrow \Phi_o=\frac{12375}{\lambda_o(A^o)}eV\)

\(\Rightarrow \lambda_o=\frac{12375}{\Phi_o(A^o)}=\frac{12375}{6.63}\, A^o=1866\, A^o\)

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