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प्रश्न
Light of wavelength 4000 Å is incident on two metals A and B. Which metal will emit photoelectrons, if their work functions are 3.8 e V and 1.6 e V respectively?
संख्यात्मक
उत्तर
Given: λ = 4000 Å = 4000 × 10-10 m
Energy of photon, E = hv = `(hc)/lambda`
`= ((6.6 xx 10^-34) xx (3.0 xx 10^8))/(4000 xx 10^-10)`
`= (4.95 xx 10^-19)/(1.6 xx 10^-19)`eV
= 3.1 eV
The incident light will yield photoelectrons if its energy is greater than the work function of the metal.
Thus, metal B will emit photoelectrons.
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