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When photons of energy 4.25eV strike the...

When photons of energy 4.25eV strike the surface of a metal A, the ejected photoelectrons have maximum kinetic energy, `T_A` (expressed in eV) and deBroglie wavelength `lambda_A`. The maximum kinetic energy of photoelectrons liberated from another metal B by photons of energy 4.20V is `T_B = T_A -1.50eV`. If the deBroglie wavelength of those photoelectrons is `lambda_B = 2lambda_A` then

A

The work function of A is 2.25 eV

B

The work function of B is 3.70 eV

C

`T_(A)=2.00 eV`

D

`T_(B)=2.75 eV`

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The correct Answer is:
D
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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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photons of energy 4.25 eV strike the surface of metal A, the ejection photoelectric have maximum kinetic energy T_(A) eV energy 4.70 eV is T_(B) = (T_(A) - 1.50) eV if the de Brogle wavelength of these photoelec tron is lambda_(B) = 2 lambda_(A) , then

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