Find the KE of the following particles that each have a de Broglie wavelength of

  • #1
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 im stuck i know that the wavelength = h/p but i dont know how to solve the others with this
 

Answers and Replies

  • #2


i found be by taking hc/ lamda squared and dividing it by the momemtum squared
 
  • #3
993
13


momentum p,
p = h/[itex]\lambda[/itex]

total energy E :
E[itex]^{2}[/itex] = p[itex]^{2}[/itex]c[itex]^{2}[/itex] + m[itex]_{o}[/itex] [itex]^{2}[/itex]c[itex]^{4}[/itex]

KE = E - m[itex]_{o}[/itex]c[itex]^{2}[/itex]
 

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