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 = (1.9xx1)}

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^2s^(-2))`

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