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`

Options

A. The work function fo (A) is `2 . 25 eV`
B. The work function of (B) is ` 3. 25 eV`
C. ` T_A 2.00 eV`
D. ` T_B = 2 . 75 eV`

Answer

D

Explanation

` u_A = h/(lambda_Am) :. U_(A)/u_(B) = ( lambda_A) = 2 ( :. lambda_B = 2 lambda_A)`
` u_B = h/(lambda_Bm)`
Now ` T_(A)/T_(B) = mu_(A^(2)) /(mu_(B^(2))) = (4)/(1)` `T_(A) = T_(B)= 1. 50`
`:. T_B =0. 50`
` T_A = 0. 50 + 1. 50 =.00 eV`
Also ` 4. 25 = hv_(0_A) + T_A ,`
` : . hv_(0_A) = 4 . 25 - 2. 00 = 2. 25 eV`
` 4. 20 = hv_(0_B) + T_B ,`
` :. hv_(0_B) = 4 . 20 - 0. 50 = 3 70 eV`.