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A free particle with initial kinetic energy 9 eV and de Broglie wavelength 1 nm enters a region of constant potential energy V0 such that the new de Broglie wavelength is now 1.5 nm. Then V0 is (A) 5 eV (B) 6 eV (C) 13.5 eV (D) 15 eV

 A free particle with initial kinetic energy 9 eV and de Broglie wavelength 1 nm enters a region
of constant potential energy V0 such that the new de Broglie wavelength is now 1.5 nm. Then
V0 is
(A) 5 eV (B) 6 eV (C) 13.5 eV (D) 15 eV

Grade:12

1 Answers

Eshan
askIITians Faculty 2095 Points
3 years ago
Dear student,

Kinetic energy of a particle with de-broglie wavelength\lambda=\dfrac{p^2}{2m}=\dfrac{h^2}{2m\lambda^2}\propto \dfrac{1}{\lambda^2}
Hence when de-broglie wavelenth increases from 1nm to 1.5nm, the kinetic energy decreases from9eVto4eV.
Hence from conservation of energy, the difference in energy must be existing in the form of potential energyV_0=5eV

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