Question

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?

Answer 1

Wavelength of ultraviolet light, λ = 2271 Å = 2271 × 10⁻¹⁰ m

Stopping potential of the metal, V₀ = 1.3 V

Planck’s constant, h = 6.6 × 10⁻³⁴ J

Charge on an electron, e = 1.6 × 10⁻¹⁹ C

Work function of the metal = `phi_0`

Frequency of light = v

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

`phi_0` = hv − eV₀

= `"hc"/lambda - "eV"_0`

= `(6.6 xx 10^(-34) xx 3 xx 10^8)/(2271 xx 10^(-10)) - 1.6 xx 10^(-19) xx 1.3` 

= `8.72 xx 10^(-19) - 2.08 xx 10^(-19)`

= `6.64 xx 10^(-19)  "J"`

= `(6.64 xx 10^(-19))/(1.6 xx 10^(-19))`

= 4.15 eV

Let v₀ be the threshold frequency of the metal.

∴ `phi_0 = "hv"_0`

`"v"_0 = phi_0/"h"`

= `(6.64 xx 10^(-19))/(6.6 xx 10^(-34))`

= 1.006 × 10¹⁵ Hz

Wavelength of red light, `lambda_"r"` = 6328 Å =  6328 × 10⁻¹⁰ m

∴ Frequency of red light, `"v"_"r" = "c"/lambda_"r"`

= `(3 xx 10^8)/(6328 xx 10^(-10))`

= 4.74 × 10¹⁴ Hz

Since v₀ > v_r, the photocell will not respond to the red light produced by the laser.