Question

A 3310 Å photon liberates an electron from a material with energy 3 × 10⁻¹⁹ J while another 5000 Å photon ejects an electron with energy 0.972 × 10⁻¹⁹ J from the same material. Determine the value of Planck’s constant and the threshold wavelength of the material.

Answer 1

The energy of ejected electron is given by E = `"hc"/λ - "hc"/λ_0`

`3 xx 10^-19 = "hc" [1/(3310 xx 10^-10) - 1/λ_0]` .......(1)

`0.972 xx 10^-19 = "hc" [1/(5000 xx 10^-10) - 1/λ_0]` .......(2)

Subtracting (2) from (1), we get

`(3 - 0.972) xx 10^-19 = "hc"/10^-10 [1/3310 - 1/5000]`

`2.028 xx 10^-19 = ("h" xx 3 xx 10^8)/10^-10 [1690/(3310 xx 5000)]`

h = `(2.028 xx 10^-19 xx 10^-19 xx 3310 xx 5000)/(3 xx 10^8 xx 1690)`

h = 6.62 × 10⁻³⁴ Js

Now W₀ = `"hc"/λ - "E"`

= `((6.62 xx 10^-34 xx 3 xx 10^8)/(3310 xx 10^-10)) - 3 xx 10^-19`

= (6 − 3) × 10⁻¹⁹

W₀ = 3 × 10⁻¹⁹ J

Threshold Wavelength,

λ₀ = `"hc"/"W"_0`

= `(6.62 xx 10^-34 xx 3 xx 10^8)/(3 xx 10^-19)`

= 6.62 × 10⁻⁷ m

λ₀ = 6620 × 10⁻¹⁰ m