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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 eV`

D

`T_(B) = 2.75 eV`

Text Solution

Verified by Experts

The correct Answer is:
A, B, C
(1) `lambda_(A) = (h)/(sqrt(2mT_(A)))` , (2) `lambda_(B) = (h)/(sqrt(2mT_(B)))`
(3) `T_(B) = T_(A) - 1.5 eV` , (4) `lambda_(B) = 2lambda_(A)`
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