Question

Find the velocity of an electron emitted by a metal whose threshold frequency is 2.24×1014 s−1 when it is exposed to visible light of wavelength 4.89×10−7 m .

Answer 1

Apply the workfunction

E = hv

h = Planck Constant = 6.626*10^-34 J s

v = speed of particle (i.e. light)

E = energy per particle J/photon

Now, get E

E = (6.626*10^-34)(2.24*10^14) = 1.484*10^-19 J/photon

Now... from the wavleenght energy:

For the wavelength:

E= h c / WL

h = Planck Constant = 6.626*10^-34 J s

c = speed of particle (i.e. light) = 3*10^8 m/s

E = energy per particle J/photon

WL = wavelength in meters

E = (6.626*10^-34)(3*10^8)/(4.89*10^-7)

E = 4.0650*10^-19 J/photon

now...

Ek = E photon - Efunction

Ek = 4.0650*10^-19 - (1.484*10^-19)

Ek = 2.581*10^-19 J

now...

Ek = 1/2*m *v^2

m electron = 9.1093*10^-31 kg

2.581*10^-19 = (1/2) * ( 9.1093*10^-31)(v^2)

v = sqrt((2.581*10^-19)*2 / ( ( 9.1093*10^-31))

v = 752777.27 m/s

v = 7.53*10^5 m/s