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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 'D'. Can you explain this answer?
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Most Upvoted Answer
An electron and a proton have the same de-Broglie wavelength. Then the...
Explanation:
De-Broglie wavelength is given by the formula:
λ = h/mv
where λ is the wavelength, h is the Planck constant, m is the mass of the particle, and v is the velocity of the particle.
Since the electron and the proton have the same de-Broglie wavelength, we can write:
h/mev(electron) = h/mpv(proton)
where me is the mass of the electron, mp is the mass of the proton, v(electron) is the velocity of the electron, and v(proton) is the velocity of the proton.
Simplifying the above equation, we get:
v(electron)/v(proton) = mp/me
Now, the kinetic energy of a particle is given by the formula:
K.E. = 1/2 mv^2
where K.E. is the kinetic energy, m is the mass of the particle, and v is the velocity of the particle.
Conclusion:
From the above equation, we can see that for the same de-Broglie wavelength, the velocity of the electron is greater than the velocity of the proton because the mass of the electron is smaller than the mass of the proton. Therefore, the kinetic energy of the electron is greater than the kinetic energy of the proton. Hence, the correct answer is option D, i.e., the kinetic energy of the electron is greater than the kinetic energy of the proton.
De-Broglie wavelength is given by the formula:
λ = h/mv
where λ is the wavelength, h is the Planck constant, m is the mass of the particle, and v is the velocity of the particle.
Since the electron and the proton have the same de-Broglie wavelength, we can write:
h/mev(electron) = h/mpv(proton)
where me is the mass of the electron, mp is the mass of the proton, v(electron) is the velocity of the electron, and v(proton) is the velocity of the proton.
Simplifying the above equation, we get:
v(electron)/v(proton) = mp/me
Now, the kinetic energy of a particle is given by the formula:
K.E. = 1/2 mv^2
where K.E. is the kinetic energy, m is the mass of the particle, and v is the velocity of the particle.
Conclusion:
From the above equation, we can see that for the same de-Broglie wavelength, the velocity of the electron is greater than the velocity of the proton because the mass of the electron is smaller than the mass of the proton. Therefore, the kinetic energy of the electron is greater than the kinetic energy of the proton. Hence, the correct answer is option D, i.e., the kinetic energy of the electron is 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...
The de broglie wavelength of both electron and proton are the same. So by the formula, the momentum(m x v) is also equal for the both. The mass of the electron is less than the mass of the proton then obviously the velocity of the electron is greater than than the velocity of proton. As the kinetic energy is the half the product of mass of the particle and square of its velocity, the kinetic energy of the electron is much more than that of proton.
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An electron and a proton have the same de-Broglie wavelength. Then the kinetic energy of the electron isa) Zero b) Infinity c) Equal to kientic energy of the proton d) Greater than the kinetic energy of proton Correct answer is 'D'. Can you explain this answer?
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