Question #a20e7

1 Answer
Jul 21, 2017

Answer:

#2.70*10^(-12)m#

Explanation:

de Brogile wavelength - #lamda=h/p=h/(v*m_e)#, where:
#lamda# = de brogile wavelength (#m#)
#h# = Planck's constant (#6m63*10^(-34)J# #s#)
#p# = momentum of the particle (#kg# #m# #s^(-1)#)
#v# = velocity of the particle (#m# #s^(-1)#)
#m_e# = electron mass = (#9.11*10^(-31) kg#)

#c~~3*10^8m# #s^(-1)#

#0.9c = 2.7*10^8m# #s^(-1)#

#lamda=(6.63*10^(-34))/((2.7*10^8)(9.11*10^(-31)))~~2.70*10^(-12)m#