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MHT CET · Physics · Dual Nature of Matter

Threshold frequency for a metal is \(15 \times 10^{14} \mathrm{~Hz}\). The light of wavelength \(6000 \)Å falls on the metal surface. Then photoelectrons
[velocity of light in air, \(c=3 \times 10^8 \mathrm{~m} / \mathrm{s}\) ]

  1. A come out with zero velocity.
  2. B come out with velocity \(3 \times 10^6 \mathrm{~m} / \mathrm{sa}\)
  3. C will not be emitted
  4. D are emitted with velocity \(c\)
Verified Solution

Answer & Solution

Correct Answer

(C) will not be emitted

Step-by-step Solution

Detailed explanation

Frequency of light of wavelength \(\lambda=6000 \mathrm{~A}^0\) is
\(f=\frac{c}{\lambda}=\frac{3 \times 10^8}{6000 \times 10^{-10}} \mathrm{~Hz}=5 \times 10^{14} \mathrm{~Hz}\)
which is less than the given threshold frequency. Hence, no photoelectric emission takes place.