Q. 94.0( 74 Votes )

# A photon of wavelength 4 × 10^{–7} 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, and (iii) the velocity of the photoelectron (1 eV= 1.6020 × 10^{–19} J).

Answer :

Given:

Wavelength, λ = 4 × 10^{–7} m

Work Function of metal = 2.13 eV

Finding Energy of Photon:

By Planck’s relation we have,

Energy, E = h×v

But we know v = [c] / [λ]

Where

c = Speed of Light

v= Frequency

λ = Wavelength

So, E = hc /λ

= [[6.626×10^{-34}] × [3×10^{8}]] / [4×10^{-7}]

= [1.9878×10^{-25}] / [4×10^{-7}]

= 4.9695×10^{-19} J

Therefore, the energy of the photon 4.9695×10^{-19} J

Finding the Kinetic Energy of Emission:

Kinetic Energy = hv - hv_{o}

= [E – W] eV

Where

E = Energy of photon in eV

W = work function of metal in eV

= {[4.9695×10^{-19}]/ [1.6020 × 10^{–19}]} – 2.13

= 3.102 – 2.13

= 0.972 eV

Therefore, the kinetic energy of emission is 0.972 eV.

Finding Velocity of photoelectron:

We know the formula for kinetic energy which is given as follows:

1/2 mv^{2} = kinetic Energy

Where

m = mass of electron

v = velocity of electron

v^{2} = [2× 0.972× 1.6020 × 10^{–19}] / [9.1× 10^{-31}]

= [3.1143× 10^{–19}] / [9.1× 10^{-31}]

v^{2} = 3.422× 10^{11}

Therefore v = √ [3.422× 10^{11}]

v = 5.849× 10^{5} m/s

Therefore, the velocity of photoelectron is 5.85× 10^{5} m/s

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NCERT - Chemistry Part-I