Answer :
1.227 A˚
V=32 volts
de-Broglie wavelength of electron λ= 2meV h
∴ λ= 2(9.1×10 −31)(1.6×10 −19 )(100) 6.6×10 −34
Or λ= 5.4×10 −10
6.6×10 −34
m⟹ λ=1.227×10 −10
m=1.227 A˚
The wavelength (λ) that is associated with an object in relation to its momentum and mass is known as de Broglie wavelength. A particle's de Broglie wavelength is usually inversely proportional to its force.The de Broglie wavelength of a particle indicates the length scale at which wave-like properties are important for that particle. De Broglie wavelength is usually represented by the symbol λ or λdB. For a particle with momentum p, the de Broglie wavelength is defined as: λdB = hp. where h is the Planck constant.
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