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Question

The minimum energy required by an electron to overcome the attractive forces of the nucleus is 0.4×1019 J. What is the maximum kinetic energy of the ejected electron if the material is radiated with a light of wavelength 3313 nm?

A
0.2×1020 J
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B
2×1020 J
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C
1×1020 J
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D
10×1020 J
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Solution

The correct option is B 2×1020 J
Step 1: Given wavelength of the incident radiation (λ)=3313 nm
Also, the minimum energy required by the electron to overcome the attractive forces ( E0) =0.4×1019 J

Step 2: 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
=3313 nm
=3313×109 m

E=6.626×1034 J.s×3×108 m/s3313×109 m
E=(6626×1033313×109×1034×3×108) J
=6×1020 J
E=0.6×1019 J

Step 3: 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,
0.6×1019 J=0.4×1019 J+K.Emax
K.Emax=(0.60.4)1019 J
=0.2×1019 J
K.Emax=2×1020 J

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