Ultraviolet light of wavelength 2271 Å from a 100 W mercury source irradiates a photo-cell made of molybdenum metal. If the stopping potential is –1.3 V, estimate the work function of the metal. How would the photo-cell respond to a high intensity (∼10⁵ W m⁻²) red light of wavelength 6328 Å produced by a He-Ne laser?

Wavelength of ultraviolet light, λ = 2271 Aº = 2271 x 10⁻¹⁰ m

Stopping potential of the metal, Vo = 1.3 V

Planck’s constant, h = 6.6 x 10⁻³⁴ J

Charge on an electron, e = 1.6 x 10^-19 C

Work function of the metal = φ_O

Frequency of light = v

We have the photo-energy relation from the photoelectric effect as:

φ_O= hv – eVo

= (6.6 x 10⁻³⁴) x ( 3 x 10⁸ ) /(2271 x 10⁻¹⁰)    – ( 1.6 x 10^-19) x 1.3

= 8.72 x 10^-19 – 2.08 x 10^-19

= 6.64 x 10^-19 J

= (6.64 x 10^-19 ) /(1.6 x 10^-19)  = 4.15 eV

Let vo be the threshold frequency of the metal .Therefore , φ_O=hv0

=>v0 = φ_O/h  = (6.64 x 10^-19)/(6.6 x 10⁻³⁴) = 1.006 x 10¹⁵ Hz

Wavelength of red light,

λr = 6328 Aº = 6328 x 10⁻¹⁰

Therefore ,the frequency of red light ,

vr = c/λr    = (3 x 10⁸) /(6328 x 10⁻¹⁰)  = 4.74 x 10¹⁴ Hz

Since vo> v_r, the photocell will not respond to the red light produced by the laser.