Question

Upon being struck by 240-nm photons, a metal ejects electrons with a maximum kinetic energy of 1.45 eV. What is the work function of this metal? (A) 3.73 eV (B) 3.13 eV (C) 4.33 eV (D) 4.92 eV

Answer 1

W = 3.73 eV

Step-by-step explanation:

Given data:

wavelength of photons 240 nm

kinetic energy 1.45 eV

B y using photoelectric equation

Kinetic Energy = Photon Energy - work friction

`K.E = (hc)/(\lambda) -W`

`(hc)/(\lambda) = (6.63* 10^(-34)* 3* 10^8)/(240* 10^(-9)* 1.6* 10^(-19)) eV`

= 5.179 eV

= 5.18 eV
`</p><p> W =  (hc)/(\lambda) - K.E`

= (5.18 - 1.45) eV

W = 3.73 eV