# A Photon of Wavelength 4 × 10–7 M Strikes on the 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 - Chemistry

Numerical

A photon of wavelength 4 × 10–7 m strikes on the metal surface, the work function of the metal being 2.13 eV. Calculate

1. the energy of the photon (eV),
2. the kinetic energy of the emission, and
3. the velocity of the photoelectron (1 eV = 1.6020 × 10–19 J).

#### Solution

(i) Energy (E) of a photon = "hν" = "hc"/lambda

Where,

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

c = velocity of light in vacuum = 3 × 10m/s

λ = wavelength of photon = 4 × 10–7 m

Substituting the values in the given expression of E:

"E" = ((6.626xx10^(-34))(3xx10^8))/(4xx10)^(-7) = 4.9695 xx 10^(-19)  "J"

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

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

= "hv" - "hv"_0

=("E" - "W")"eV"

= ((4.9695xx10^(-19))/(1.6020xx10^(-19))) "eV" - 2.13 " eV"

= (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,

1/2 "mv"^2 = "hv" - "hv"_0

=> "v" = sqrt((2("hv" - "hv"_0))/"m")

where ("hv" - "hv"_0) 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" = sqrt((2xx(0.9720xx1.6020xx10^(-19))"J")/(9.10939xx10^(-31)"kg"

= sqrt(0.3418 xx 10^12 "m"^2"s"^(-2))

v = 5.84 × 105 ms–1

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

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Chapter 2: Structure of Atom - EXERCISES [Page 69]

#### APPEARS IN

NCERT Chemistry Part 1 and 2 Class 11
Chapter 2 Structure of Atom
EXERCISES | Q 2.9 | Page 69

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