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Question: Answered & Verified by Expert
If the photon of wavelength 150 pm strikes an atom and one of its inner bound electron is ejected out with a velocity of 1.5×107 m/s, what is the energy with which it is bound to the nucleus?
ChemistryStructure of AtomBITSATBITSAT 2019
Options:
  • A 1.2×102 eV
  • B 2.15×103 eV
  • C 7.6×103 eV
  • D 8.12×103 eV
Solution:
1644 Upvotes Verified Answer
The correct answer is: 7.6×103 eV

Total energy E of any photon is given by the relation : E=hcλ

Where, h= Planck's constant =6.6×10-34 Js

c= Velocity of light =3×108 m/s

λ= wavelength =1.5×10-10 m, i.e, 150 pm

Thus, E=hcλ=6.6×10-34×3×1081.5×10-10; E=1.32×10-15 J and, energy of ejected electron E'

E'=12mv2,

where,

m= mass of electron =9.1×10-31 kg

v= velocity of electron

=1.5×107 m/s

E'=12mv2=12×9.1×10-31×1.5×1072

E'=1.024×10-16

Thus, total energy of photon = Binding energy of electron

B + energy of ejected electron E'

Thus, 1.32×10-15=B+E

 B=E-E'=1.32×10-15-1.024×10-16

=1.2176×10-15 J

=1.2176×10-151.6×10-19 eV=7.6×103 eV

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