1.

Velocity of helium atom at 300K is 2.40 xx 10^2 meter per second. What is its wavelength? (He = 4)

Answer»

0.416 nm
0.83 nm
803 Å
8000 Å

Solution :`LAMBDA = (6.625 XX 10^(-34))/(4xx 67 xx 10^(-27) xx 2.4xx10^2)= 0.416 nm`


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