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Ratio of the work functions of two metal...

Ratio of the work functions of two metal surface is `1 : 2`. If threshold wavelength of photoelectric effect for the 1st metal is `6000Å`, what is the corresponding value for the 2nd metal surface?

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If work function of 1st and 2nd metals be `W_(0) and W'_(0)`, respectively then,
`(W_(0))/(W'_(0)) = (hf_(0))/(hf'_(0)) = (hc//lamda_(0))/(hc//lamda'_(0)) = (lamda'_(0))/(lamda_(0))`
or, `lamda'_(0) = lamda_(0) xx (W_(0))/(W'_(0)) = 6000 xx (1)/(2) = 3000 Å`
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