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Question
When electromagnetic radiatin of wavelength 300 nm falls on the surface of sodium, electrons are emitted with a kinetic energy of 1.68×105Jmol−1. What is the minimum energy needed to remove an electron from sodium? What is the maximum wavelength that will cause a photoelectron to be emitted?
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Solution
Given:
Wavelength of electromagnetic radiationλ=300 nm=300×10−9 m
Kinetic energy of electromagnetic radiation K.E.=1.68×105 J mol−1
Step-1: Calculation for the energy of one mole photon having wavelength 300 nm.
The energy of photon or electromagnetic radiation is given by:
E=hv=hc/λ
where,h→planck's constant=6.626×10−14J sec
v→frequency of electromagnetic radiation
c→velocity of light=3×108 ms−1
λ→wavelength of electromagnetic radiation=300 nm=300×10−9 m
Substituting the values, we get
E=6.626×10−34J sec×3×108 msec−1300×10−9m
=0.06626×10−34+8+9J
E=6.626×10−19J
Now, the energy of the one mole of photon
E=6.626×10−19(J)×6.022×1023(mol−1)
E=3.99×105 J mol−1
Step-2: Calculation for maximum wavelength that will cause a photoelectron to be emitted.
Work function: The minimum amount of the energy needed to remove an e− from the surface of a given solid.
∵ We know that; Energy of photon = K.E. + work function
Work function(hv0)=Energy of photon - Kinetic energy
(6.626×10−19−1.68×105×6.022×1023)J
=(6.626×10−19−2.79×10−19)J
=3.84×10−19J
Now corresponding wavelength
E=hcλ=3.84×10−19J
λ=hc3.84×10−19J=6.626×10−34Js×3×108 ms−13.84×10−19J
λ=19.878×10−34+83.84×10−19m=5.176×10−2610−19
λ=5.176×10−7m
λ=517.6 nm
Final answer:
(i) Minimum amount of energy needed to remove an e− from sodium =3.84×10−19J
(ii) Maximum wavelength that will cause a photoelectron to be emitted =517.6 nm
Wavelength of electromagnetic radiationλ=300 nm=300×10−9 m
Kinetic energy of electromagnetic radiation K.E.=1.68×105 J mol−1
Step-1: Calculation for the energy of one mole photon having wavelength 300 nm.
The energy of photon or electromagnetic radiation is given by:
E=hv=hc/λ
where,h→planck's constant=6.626×10−14J sec
v→frequency of electromagnetic radiation
c→velocity of light=3×108 ms−1
λ→wavelength of electromagnetic radiation=300 nm=300×10−9 m
Substituting the values, we get
E=6.626×10−34J sec×3×108 msec−1300×10−9m
=0.06626×10−34+8+9J
E=6.626×10−19J
Now, the energy of the one mole of photon
E=6.626×10−19(J)×6.022×1023(mol−1)
E=3.99×105 J mol−1
Step-2: Calculation for maximum wavelength that will cause a photoelectron to be emitted.
Work function: The minimum amount of the energy needed to remove an e− from the surface of a given solid.
∵ We know that; Energy of photon = K.E. + work function
Work function(hv0)=Energy of photon - Kinetic energy
(6.626×10−19−1.68×105×6.022×1023)J
=(6.626×10−19−2.79×10−19)J
=3.84×10−19J
Now corresponding wavelength
E=hcλ=3.84×10−19J
λ=hc3.84×10−19J=6.626×10−34Js×3×108 ms−13.84×10−19J
λ=19.878×10−34+83.84×10−19m=5.176×10−2610−19
λ=5.176×10−7m
λ=517.6 nm
Final answer:
(i) Minimum amount of energy needed to remove an e− from sodium =3.84×10−19J
(ii) Maximum wavelength that will cause a photoelectron to be emitted =517.6 nm
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