A photon of wavelength 4.0 x m strikes on metal surface, the work function of the metal being 2.13 eV. Calculate:

i. The energy of the photon (eV).

ii. The kinetic energy of the emission.

Iii. The velocity of the photoelectron.

(i) apply planck's energy equation, E = hc/λ 

(ii) K.E = E - (work function)

(iii) 1/2 m= K.E

do calculation yourself to understand, 

(1) Energy of the photon (E) = hν = hc/λ = (6.626×10^-34 Js×3.0×10⁸ ms^-1)/4×10^-7 m = 4.97×10^-19 J
= 4.97×10^-19/1.602×10^-19 eV

(2) Kinetic energy of emission (1/2 mv²) = hν- hνo = 3.10-2.13 = 0.97 eV

(3) 1/2 mv² = 0.97 eV = 0.97×1.602×10^-19 J
⇒ 1/2×(9.11×10^{-31 kg})×v² = 0.97×1.602×10^-19 J
⇒ v² = 0.341×10¹² = 34.1×10¹⁰
⇒ v = 5.84×10⁵ ms^-1

First of all we have to find the enenry(E) of Photon by using the Formula

E=hc/λ

here 'h' is the Plank's constant. 

'c' is the Speed of light

and λ is the wavelength

Now we have to think what is the term "WORK FUNCTION" used in the Question

WORK FUNCTION is the MINIMUM amount of energy required to release a ELECTRON from the Metal

It is given in the Question

Now we are able to give the answer of the 2nd part of the Question

i.e Kinetic energy of emission== Total energy of photon- Energy required for emission(work function)

Now the question is WHAT IS THE VELOCITY OF PHOTOELECTRON??

From 2nd part of Question We know the Kinetic Energy of Photoelectron.

Use the formula,

K.E= ½*mass*(velocity)²

for the value of "mass of electron" , "plank's constant" and "velocity of light "use your book.

THANK YOU

(1)We know λ = 4 × 10^–7 m(given)

C = 3 x 108

From the equation E= hv or hc/ λ

Where, h = Planck’s constant = 6.626 × 10^–34 Js

c = velocity of light in vacuum = 3 × 10⁸ m/s

λ = wavelength of photon = 4 × 10^–7 m

Substituting the values in the given expression of E:

Hence, the energy of the photon is 4.97 × 10^–19 J.

(ii) The kinetic energy of emission Ek is given by

= (3.1020 – 2.13) eV

= 0.9720 eV

Hence, the kinetic energy of emission is 0.97 eV.

(iii) The velocity of a photoelectron (ν) can be calculated by the expression,

Where, (hv-hv₀) is the kinetic energy of emission in Joules and ‘m’ is the mass of the photoelectron. Substituting the values in the given expression of v:

v = 5.84 × 10⁵ ms^–1

Hence, the velocity of the photoelectron is 5.84 × 10⁵ ms^–1.

Energy of photon= hc/λ

h=plank constant = 6.63*

c=speed of light=3*

h*c= 1240 eV

λ= wavelength of light

E= / joule

E=3.102 eV

2.)Kinetic energy of emission= energy of photon- work function

                                      =3.102-2.13 eV

                                      = 0.972 eV

3.)Velocity of free Electrons 

kinetic energy of electron = 1/2*m* = 0.972 eV

                                                   v=

                                                      =5.84 m/s