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Sahay Sir > Question Answers > Uncategorised (JEE Advanced Physics by BM Sharma + GMP Solutions) > Two identical metal plates shown photoelectric effect. Light of wavelength lemda A falls on plate A and lemda B falls on plate B, lemda A = 2 lemda B. The maximum KE of the photoelectrons are

Two identical metal plates shown photoelectric effect. Light of wavelength lemda A falls on plate A and lemda B falls on plate B, lemda A = 2 lemda B. The maximum KE of the photoelectrons are

lemda A = 2 lemda B. The maximum KE of the photoelectrons are Two identical metal plates shown photoelectric effect. Light of wavelength lemda A falls on plate A and lemda B falls on plate B