Question

Physics. Urgent

The work needed to remove an electron from the surface of rutile TiO₂ is 3.0 eV. What is the longest wavelength at which we can have free electrons within the material and outside of it? What will be the stopping potential for electrons if TiO₂ is irradiated with light with wavelength of 200 nm? What will the stopping potential be if TiO₂ is irradiated with 530 nm wavelength? What is the photoelectron energy ratio for aforementioned wavelengths?

Answer 1

given

workfucntion, Wo = 3.0 eV

to liberate electron from the surface, energy of photon must be atleast 3.0 eV.

let lamda_max is the logest wavelength for which electron is emited from the surface.

minimum energy of photon, E_min = 3.0 eV

= 3.0*1.6*10^-19 J

now use, E_min = h*c/lamda_max

==> lamda_max = h*c/E_min

= 6.626*10^-34*3*10^8/(3*1.6*10^-19)

= 4.14*10^-7 m (or) 414 nm <<<<<<<<<---------------Answer

when lamda = 200 nm,

maximum kinetic energy of electron,

KE_max = E_photon - Wo

= h*c/lamda - Wo

= 6.626*10^-34*3*10^8/(200*10^-9) - 3*1.6*10^-19

KE_max = 5.14*10^-19 J

If Vo is the stoppiong potential.

use, KE_max = q*Vo

==> Vo = KE_max/q

= KE_max/q

= 5.14*10^-19/(1.6*10^-19)

= 3.21 V <<<<<<<<<<<<<-----------------------------Answer

when lamda = 530 nm,

energy of photon, E_photo = h*c/lamda - Wo

= 6.626*10^-34*3*10^8/(530*10^-9)

= 3.75*10^-19 J

= 3.75*10^-19/(1.6*10^-19)

= 2.34 eV

clearly E_photon < Wo

so, electron is not emitted. <<<<<<<<<<<<<-----------------------------Answer

Please comment for further clarification.