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When a metal surface is illuminated by a monochromatic light of wave-length `lambda`, then the potential difference required to stop the ejection of electrons is `3V`. When the same surface is illuminated by the light of wavelength `2lambda`, then the potential difference required to stop the ejection of electrons is `V`. Then for photoelectric effect, the threshold wavelength for the metal surface will beA. `6lambda`B. `4lambda//3`C. `4lambda`D. `8lambda` |
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Answer» Correct Answer - C `(hc)/lambda=(hc)/lambda_(0)+eV_(0)` or `hc(1/lambda-1/lambda_(0))=eV_(0)` |
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