Class 11 Chemistry - Chapter Structure of Atom NCERT Solutions | A photon of wavelength 4 × 10&ndas
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).
(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.
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5 Comment(s) on this Question
best useful material for jee mains & advanced lerners
please could you mind to help me with equations of 3000A and change to be 3.0*10m = f=3.0*10/3.0*10 just now
Caluclate the de-broglie wavelength for metal surface electron which have the work function 30joule and wavelength of photon which is provide to the metal surface is 10 rase power -6m
Yes their is some typing mistake the answer comes out to be 1.848988 * 10 to the power 4
I have a doubt because-: v2 =2Ã.97Ã1.6Ã10-19/9.1Ã10-31 When I solve this I got answer as 341Ã10power9 and after underroot I got answer as-:18.466Ã10power3 Please reply as soon as possible.
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