What is the de broglie wavelength of an electron traveling at 1.34×105 m/s ?

To solve this problem, we make use of the formula:

λ = h/mv 

where:

λ = wavelength of electron

m = mass of electron = 9.11e-31 kg

v = speed of electron = 1.34×10^5 m/s

h = constant

Hence,
λ = (6.626x10^-34 J-s) / [(9.11e-31 kg)(1.34 x 10^5 m/s)] 
λ = 5.43x10^-9 m or 5.43 nm

answer: the correct answer id option(d).

explanation:

snell's law refraction:

the angle of refraction =

the angle of incidence =

refractive index of air = = 1

refractive index of glycerol = 1.47

from the given options, the most closest answer to 63.72° is 65° of option (d).

the higher the current the higher the mag field around it the higher the influence on the compass