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. Which one of the following statements is correct? [velocity of light, \(\left.\mathrm{c}=3 \times 10^{8} \mathrm{~m} / \mathrm{s}\right]\)
- A photoelectrons are emitted with velocity c.
- B photoelectrons come out with velocity \(3 \times 10^{6} \mathrm{~m} / \mathrm{s}\)
- C photoelectrons come out with zero velocity.
- D photoelectrons will not be emitted.
Answer & Solution
Correct Answer
(D) photoelectrons will not be emitted.
Step-by-step Solution
Detailed explanation
\(v_{0} =15 \times 10^{14} \mathrm{~Hz} \)
\( \therefore \lambda_{0}=\frac{C}{v} =\frac{3 \times 10^{8}}{15 \times 10^{14}} \)
\( =2 \times 10^{-7} \mathrm{~m} \)
\( =2000\)Å
\( \lambda =6000\)Å
Since \(\lambda>\lambda_{0}\), photo electrons will not be emitted
\( \therefore \lambda_{0}=\frac{C}{v} =\frac{3 \times 10^{8}}{15 \times 10^{14}} \)
\( =2 \times 10^{-7} \mathrm{~m} \)
\( =2000\)Å
\( \lambda =6000\)Å
Since \(\lambda>\lambda_{0}\), photo electrons will not be emitted
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