A molecule of oxygen gas (O2 ) has a velocity of 2800 m / s. What is its de Broglie wavelength?

1 Answer
Truong-Son N.
Feb 6, 2018

I got #"4.454 pm"#.

The de Broglie wavelength of a mass-ive particle is given by

#lambda = h/(mv)#,

with #h = 6.626 xx 10^(-34) "J"cdot"s"# being Planck's constant, #m# the mass in #"kg"#, and #v# the velocity in #"m/s"#.

We'll need its molecular mass.

#"0.031998 kg"/cancel"mol" xx cancel"1 mol"/(6.0221413 xx 10^(23)) = 5.313 xx 10^(-26) "kg"#

Therefore,

#color(blue)lambda = (6.626 xx 10^(-34) cancel"kg"cdot"m"^cancel(2)"/"cancel"s")/(5.313 xx 10^(-26) cancel"kg" cdot 2800 cancel"m/s")#

#= 4.454 xx 10^(-12) "m"#

#=# #color(blue)"4.454 pm"#

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