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An electron and a proton have the same de-Broglie wavelength. Then the kinetic energy of the electron is
- a)Zero
- b)Infinity
- c)Equal to kientic energy of the proton
- d)Greater than the kinetic energy of proton
Correct answer is option 'D'. Can you explain this answer?
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An electron and a proton have the same de-Broglie wavelength. Then the...
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An electron and a proton have the same de-Broglie wavelength. Then the...
Explanation:
De-Broglie Wavelength:
- The de-Broglie wavelength of a particle is given by the formula: λ = h / p, where λ is the wavelength, h is the Planck's constant, and p is the momentum of the particle.
Given:
- Let λ(e) be the de-Broglie wavelength of the electron and λ(p) be the de-Broglie wavelength of the proton.
- Given that λ(e) = λ(p).
Kinetic Energy and de-Broglie Wavelength:
- The kinetic energy of a particle is related to its momentum by the formula: KE = p^2 / 2m, where KE is the kinetic energy and m is the mass of the particle.
- As the de-Broglie wavelength of the electron and the proton is the same, their momenta are also the same, i.e., p(e) = p(p).
Comparison of Kinetic Energies:
- The kinetic energy of the electron (KE(e)) is given by KE(e) = p(e)^2 / 2m(e).
- The kinetic energy of the proton (KE(p)) is given by KE(p) = p(p)^2 / 2m(p).
- Since p(e) = p(p), we can compare the kinetic energies: KE(e) = p^2 / 2m(e) and KE(p) = p^2 / 2m(p).
Conclusion:
- Since the mass of the electron (m(e)) is less than the mass of the proton (m(p)), the kinetic energy of the electron (KE(e)) will be greater than the kinetic energy of the proton (KE(p)).
- Therefore, the correct answer is option 'D': Greater than the kinetic energy of the proton.
De-Broglie Wavelength:
- The de-Broglie wavelength of a particle is given by the formula: λ = h / p, where λ is the wavelength, h is the Planck's constant, and p is the momentum of the particle.
Given:
- Let λ(e) be the de-Broglie wavelength of the electron and λ(p) be the de-Broglie wavelength of the proton.
- Given that λ(e) = λ(p).
Kinetic Energy and de-Broglie Wavelength:
- The kinetic energy of a particle is related to its momentum by the formula: KE = p^2 / 2m, where KE is the kinetic energy and m is the mass of the particle.
- As the de-Broglie wavelength of the electron and the proton is the same, their momenta are also the same, i.e., p(e) = p(p).
Comparison of Kinetic Energies:
- The kinetic energy of the electron (KE(e)) is given by KE(e) = p(e)^2 / 2m(e).
- The kinetic energy of the proton (KE(p)) is given by KE(p) = p(p)^2 / 2m(p).
- Since p(e) = p(p), we can compare the kinetic energies: KE(e) = p^2 / 2m(e) and KE(p) = p^2 / 2m(p).
Conclusion:
- Since the mass of the electron (m(e)) is less than the mass of the proton (m(p)), the kinetic energy of the electron (KE(e)) will be greater than the kinetic energy of the proton (KE(p)).
- Therefore, the correct answer is option 'D': Greater than the kinetic energy of the proton.
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An electron and a proton have the same de-Broglie wavelength. Then the kinetic energy of the electron isa)Zerob)Infinityc)Equal to kientic energy of the protond)Greater than the kinetic energy of protonCorrect answer is option 'D'. Can you explain this answer?
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