Atomic StructureEasy
Question
The de-Broglie wavelength of an electron in the ground state of hydrogen atom is :
[K.E. = 13.6eV; leV = 602 × 10-19J]
[K.E. = 13.6eV; leV = 602 × 10-19J]
Options
A.33.28 nm
B.3.328 nm
C.0.3328 nm
D.0.0332 nm
Solution
We know that
K.E. =
mv2
∴ v =
=
2.18824 × 106m/s
No since, λ =
= 0.3328 × 10-9
= 0.3328 nm
K.E. =
mv2∴ v =
=
2.18824 × 106m/s
No since, λ =
= 0.3328 × 10-9
= 0.3328 nm
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