Question

Barium has a work function of 2.48 eV.

A) What is the maximum kinetic energy of electrons if the metal is illuminated by UV light of wavelength 335

nm ? (answer in eV)

B) What is their speed? Please include appropriate units.

Please show all work Thank you.

Answer 1

Answer 2

Max KE = hf - work function = hc/lamba - work function = 3.71-2.48 eV =1.2232 eV

KE = .5MV^2

V = sqrt(2KE/M) = 655940.9818 m/sec

Answer 3

A0energy=hc/wavelength

hc=1.23 eV*micro metre

energy=1.23*10^3/335=3.7ev

max kinetic energy=3.7-2.48=1.22eV

B)v=sqrt(2*kE/m)=65.49*10^4 m/s

Answer 4

A
E = Wo + KE

h*c/lamda = 2.48 eV + KE

6.626*10^-34*3*10^8/335*10^-9 = 2.48 eV + KE

5.93*10^-19 J = 2.48 eV + KE

3.7 eV = 2.48 eV + KE

KE = 3.7 - 2.48

= 1.22 eV

B) KE = 0.5*m*v^2

v = sqrt(2*KE/m)

= sqrt(2*1.22*1.6*10^-19/9.1*10^-31)

= 6.55*10^5 m/s