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, \(c=3 \times 10^8 \mathrm{~m} / \mathrm{s}\) ]
- A come out with zero velocity.
- B come out with velocity \(3 \times 10^6 \mathrm{~m} / \mathrm{sa}\)
- C will not be emitted
- D are emitted with velocity \(c\)
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.
\(f=\frac{c}{\lambda}=\frac{3 \times 10^8}{6000 \times 10^{-10}} \mathrm{~Hz}=5 \times 10^{14} \mathrm{~Hz}\)
which is less than the given threshold frequency. Hence, no photoelectric emission takes place.
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