Single Choice

$$A$$ and $$B$$ are two metals with threshold frequencies $$1.8\times 10^{14}\ $$$$Hz$$ and $$2.2\times 10^{14}\ $$$$Hz$$. Two identical photons of energy $$0.825\ $$$$eV$$ each are incident on them. Then photoelectrons are emitted by: (Taking $$h=6.626\times 10^{-34}\ $$$$Js$$)

A$$B$$ alone
B$$A$$ alone
Correct Answer
CNeither $$A$$ nor $$B$$
DBoth $$A$$ and $$B$$

Solution

Frequency of photon corresponding to energy $$0.825\ eV =\dfrac{0.825\times 1.6\times 10^{-19}}{6.626\times 10^{-34}}=1.99\times 10^{14}\ Hz$$
Since, this frequency is greater than the frequency required for emission of photoelectrons from metal $$A$$, photoelectrons are emitted from metal $$A$$ but not from metal $$B$$.


SIMILAR QUESTIONS

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In photoelectric effect, the number of photo-electrons emitted is proportional to:

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A certain metallic surface is illuminated with monochromatic light of wavelength, $$\lambda$$. The stopping potential for photo-electric current for this light is $$3V_0$$. If the same surface is illuminated with light of wavelength $$2\lambda$$, the stopping potential is $$V_0$$. The threshold wavelength for this surface for photo-electric effect is:

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The photoelectric work function for a metal is 6×10−19J. Threshold wavelength is equal to _______________.

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