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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 metal A is 2.25 eV

B

The work function of metal B is 4.20 eV

C

`(KE)_A=2eV`

D

`(KE)_B=2.75 eV`

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When photon of energy 25eV strike the surface of a metal A, the ejected photelectron have the maximum kinetic energy photoelectrons have the maximum kinetic energy T_(A)eV and de Brogle wavelength lambda_(A) .The another kinetic energy of photoelectrons liberated from another metal B by photons of energy 4.76 eV is T_(B) = (T_(A) = 1.50) eV .If the de broglie wavelength of these photoelectrons is lambda_(B) = 2 lambda_(A) then i. (W_(B))_(A) = 2.25 eV II. (W_(0))_(B) = 4.2 eV III T_(A) = 2.0 eV IV. T_(B) = 3.5 eV

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