COMEDK · Physics · 27. Dual Nature of Matter
The threshold frequency for a metal surface is '\(n_0\)'. A photo electric current '\(I\)' is produced when it is exposed to a light of frequency \(\left(\dfrac{11}{6}\right) \mathrm{n}_{\mathrm{o}}\) and intensity \(\mathrm{I}_{\mathrm{n}}\). If both the frequency and intensity are halved, the new photoelectric current '\(\mathrm{I}^1\)' will become:
- A \(\mathrm{I}^1=2 \mathrm{I}\)
- B \(\mathrm{I^1=0}\)
- C \(\mathrm{I}^1=\dfrac{1}{2} \mathrm{I}\)
- D \(\mathrm{I}^1=\dfrac{1}{4} \mathrm{I}\)
Answer & Solution
Correct Answer
(B) \(\mathrm{I^1=0}\)
Step-by-step Solution
Detailed explanation
The photoelectric effect occurs only when the incident frequency \(\nu\) is greater than or equal to the threshold frequency \(\nu_0\). Initially, the incident frequency is \(\nu_1 = \dfrac{11}{6}\nu_0\). Since \(\nu_1 > \nu_0\), photoelectric emission occurs, and the current is…
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