What is the kinetic energy of particles with a de Broglie wavelength of 0.50 nm?

AI Thread Summary
The discussion focuses on calculating the kinetic energy of various particles with a de Broglie wavelength of 0.50 nm. For photons, the kinetic energy is determined using the formula E = hc/λ, resulting in 2480 eV. The participants express difficulty in applying the de Broglie wavelength formula, p = h/λ, to find the kinetic energies of electrons, neutrons, and α particles. They mention the total energy equation E² = p²c² + m₀²c⁴ and the relationship KE = E - m₀c² for further calculations. The conversation highlights the need for clarity in applying quantum mechanics principles to different particle types.
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Find the kinetic energy of the following particles that each have a de Broglie wavelength of 0.50 nm.?

(a) photons
____eV = 2480eV
(b) electrons
____eV
(c) neutrons
____eV
(d) α particles
____eV

i know E= hc/wavelength which = a and after that I am stuck i know that the wavelength = h/p but i don't know how to solve the others with this
 
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i found be by taking hc/ lamda squared and dividing it by the momentum squared
 


momentum p,
p = h/\lambda

total energy E :
E^{2} = p^{2}c^{2} + m_{o} ^{2}c^{4}

KE = E - m_{o}c^{2}
 
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