Question No:1If the kinetic energy of electron in hydrogenatom is 3.4e

1.	Question No:1If the kinetic energy of electron in hydrogenatom is 3.4eV, then find its de-Broglieswavelength2 = 2112 = 21 x 0.53 Å2 = 27 x 0.53 x 2Å1 - 124003.4OOONext Question
Answer» 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 MARK ME AS BRAINLIEST

Question No:1If the kinetic energy of electron in hydrogenatom is 3.4eV, then find its de-Broglieswavelength2 = 2112 = 21 x 0.53 Å2 = 27 x 0.53 x 2Å1 - 124003.4OOONext Question

Answer»

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