Question

The work function for cesium is 1.9eV.

A) Find the threshold frequency for the photoelectric effect.

B) Find the stopping potential if the wavelength of the incident light is 400nm.

Answer 1

The work function for cesium is W = 1.9eV.

                                                  = 1.9 x 1.6 x 10 ^-19 J

A) The threshold frequency for the photoelectric effect f = W / h

Where h = planck's constant = 6.625 x 10 ^-34 J s

Substitute values you get f = 4.588 x10 ¹⁴ Hz

B) The stopping potential if the wavelength of the incident light is 400nm is V = ?

V e= (hc / $\lambda$ ) - W

Where $\lambda$ = 400 x 10 ^-9 m

Substitute values you get V e= (4.96875 x 10 ^-19 ) - (1.9 x1.6 x 10 ^-19 )

where e = 1.6 x 10 ^-19 C

So, V = 3.105 - 1.9

        = 1.205 volts