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Question

The minimum energy required by an electron to overcome the attractive forces of the nucleus is 0.6×1019 J. What is the maximum kinetic energy of the ejected electron if the material is radiated with a light of wavelength 2210 nm?

A
10×1020 J
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B
5×1020 J
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C
3×1020 J
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D
9×1020 J
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Solution

The correct option is C 3×1020 J
Given : wavelength of the incident radiation (λ)=2210 nm
Also, the minimum energy required by the electron to overcome the attractive forces ( E0) =0.6×1019 J

Finding the energy of the incident radiation (E)
We know that, E=hcλ
Where, h= Planck's constant =6.626×1034 Js
c= speed of light =3×108 m/s
λ= Wavelength of the incident radiation
=2210 nm
=2210×109 m

E=6.626×1034 J.s×3×108 m/s2210×109 m
E=(6626×1032210×109×1034×3×108) J
=9×1020 J
E=0.9×1019 J

According to Einstein's photoelectric equation,
E=E0+K.Emax
Where, E is the energy of the incident light
E0 is the work function.
KEmax is the maximum kinetic energy of the ejected electron.
Substituting the values of E and E0,
0.9×1019 J=0.6×1019 J+K.Emax
K.Emax=(0.90.6)1019 J
=0.3×1019 J
K.Emax=3×1020 J

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