When photons of energy 4.25 eV strikes the surface of a metal ‘A’, the ejected photoelectrons have m

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When photons of energy 4.25 eV strikes the surface of a metal ‘A’, the ejected photoelectrons have maximum kinetic energy TA (in eV) and de Broglie wavelength λA. The maximum kinetic energy of photoelectrons liberated from another metal ‘B’ by photons of energy 4.20 eV is TB (= TA – 1.50 eV). If the de Broglie wave length of these photoelectrons is λB (= 2λA), then

Accepted Answer

$hv = \phi + (K.E.)_{\max}$For A: $\phi_{A} + T_{A}	ext{ and }\lambda_{A} = \frac{h}{\sqrt{2mT_{A}}}$For B: $4.20 = \phi_{B} + T_{B}	ext{ and }\lambda_{B} = \frac{h}{\sqrt{2mT_{B}}}$As TB = TA – 1.50 and λB = 2 λA$\phi_{A}$= 2.25 eV; $\phi_{B}$= 3.70 eV; TA = 2.0 eV; TB = 0.5 eV

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