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two metals A and b have work function 2eV and 4eV respectively which of the metals has a smaller threshold wavelenght? pls solve ftafat

Let us consider a particle of mass ‘m’ rotating about a

We know,

E = hν = hc/λ

Thus, λ or threshold wavelength is inversely proportional to the energy or the work-function.

So, metal B which has higher work function of 4 eV will have smaller threshold wavelength.

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Wo = hvo so, vo for b is more consequently threshold wavelength for it is less....

since, wavelength = (wavelength)/(frequency)

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small threshold wavelength means greater threshold frequency so metal whose work function is lower has more threshold frequency and so has smaller wavelength and so metal a has smaller threshold wavelength

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We hav a formula work function Wo= hfo= hc/wavelength...........this shows work function is inversly proportional to wavelength.......so metal B has smaller wave length

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