Question

Question No:1
If the kinetic energy of electron in hydrogen
atom is 3.4eV, then find its de-Broglies
wavelength
2 = 211
2 = 21 x 0.53 Å
2 = 27 x 0.53 x 2Å
1 - 12400
3.4
OOO
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Answer 1

Answer:

2.17 angstrom

Explanation:

wavelength of electron is given by

=> h

√( 2m KE)

apply this

we get

6.626 × 10^-(34)

√[ 2 × 9.1 × 10^(-31) × 3.4 × 1.6 × 10^-16]

{ 1 ev = 1.6 × 10^ -16 }

we get

lamda => 2.17 angstrom

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