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When photon of energy 4.25 eV strike the...

When photon of energy 4.25 eV strike the surface of metal A, the ejected photoelectrons have maximum kinetic energy `T_A` and de Broglie wavlength `lamda_A`. The maximum kinetic energy of photoelectrons liberated from another metal B by photons of energy 4.70 eV is `T_B=(T_A-1.50)`eV. If the de Broglie wavelength of these photoelectrons is `lamda_B=2lamda_A`, then

A

the work function of A is 2.25 eV

B

the work function of B is 4.20 eV

C

`T_(A)`=2.00 V

D

`T_(B)` =2.75 eV

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The correct Answer is:
A, B, C
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