Question

The work function for caesium atom is 1.9 eV. Calculate

1. the threshold wavelength and
2. the threshold frequency of the radiation. If the caesium element is irradiated with a wavelength 500 nm, calculate the kinetic energy and the velocity of the ejected photoelectron.

Answer 1

It is given that the work function (W₀) for caesium atom is 1.9 eV.

(a) From the expression, `"W"_0 = ("hc")/lambda_0`, we get

`lambda_0 = ("hc")/"W"_0`

Where,

λ₀ = threshold wavelength

h = Planck’s constant

c = velocity of radiation

Substituting the values in the given expression of (λ₀):

`lambda_0 = ((6.626 xx 10^(-34)" Js")(3.0xx10^8 " ms"^(-1)))/(1.9xx1.602 xx 10^(-19) " J")`

`lambda_0 = 6.53 xx 10^(-7) " m"`

Hence, the threshold wavelength `lambda_0` is 653 nm.

(b) From the expression, `"W"_0 = "hv"_0` we get:

`"v"_0 = "W"_0/"h"` 

Where

ν₀ = threshold frequency

h = Planck’s constant

Substituting the values in the given expression of ν_0:

`"v"_0 = (1.9 xx 1)`

ν₀ = threshold frequency

h = Planck’s constant

Substituting the values in the given expression of ν_0:

`"v"_0 = (1.9xx1.602xx10^(-19)" J")/(6.626xx10^(-34)" Js")`

(1 eV = 1.602 × 10^–19 J)

ν₀ = 4.593 × 10¹⁴ s^–1

Hence, the threshold frequency of radiation (ν₀) is 4.593 × 10¹⁴ s^–1.

(c) According to the question:

Wavelength used in irradiation (λ) = 500 nm

Kinetic energy = h (ν – ν₀)

`= "hc"(1/lambda - 1/lambda_0)`

`= (6.626 xx 10^(-34) " Js") (3.0 xx 10^8 " ms"^(-1)) ((lambda_0 - lambda)/(lambdalambda_0))`

`= (1.9878 xx 10^(-26) " Jm") [((653 - 500)10^(-9)" m")/((653)(500)10^(-18)" m"^2)]`

`= ((1.9878 xx 10^(-26))(153xx10^9))/((653)(500))`J

= 9.3149 × 10^–20 J

Kinetic energy of the ejected photoelectron = 9.3149 × 10^–20J

Since `"K.E" = 1/2 " mv"^2 = 9.3149 xx 10^(-20) " J"`

`"v" = sqrt((2(9.3149 xx 10^(-20) " J"))/(9.10939 xx 10^(-31) " kg"))`

`= sqrt(2.0451 xx 10^11 " m"^2" s"^(-2))`

Hence, the velocity of the ejected photoelectron (v) is 4.52 × 10⁵ ms^–1