The mass of an electron is 9.1 x10 31 kg if its k.e. is 3 x 10 25 j, calculate its wavelength

The mass of an electron is 9.1 × 10–31 kg. If its K.E. is 3.0 × 10–25 J, calculate its wavelength.

From de Broglie’s equation,

`lambda = "h"/("mv")`

Given,

Kinetic energy (K.E) of the electron = 3.0 × 10-25 J

Since `"K.E" = 1/2 "mv"^2`

:. Velocity (v) = `sqrt((2"K.E")/"m")`

`= sqrt((2(3.0xx10^(-25) " J"))/(9.10939 xx10^(-31)))`

`= sqrt(6.5866xx10^4)`

`v = 811.579 " ms"^(-1)`

Substituting the value in the expression of λ:

`lambda = (6.626 xx 10^(-34) " Js")/((9.10939xx10^(-31) " kg")(811.579 " ms"^(-1)))`

`lambda = 8.9625 xx 10^(-7) " m"`

Hence, the wavelength of the electron is 8.9625 × 10–7 m.

Concept: Towards Quantum Mechanical Model of the Atom - Dual Behaviour of Matter: De Broglie's relationship

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Text Solution

Solution : from de Broglie's equation <br> `lambda= h/(mv)` <br> given <br> kinetic energy (K.E) of the electron =`3.0 xx 10^(-25)J` <br> since ` K.E =1/2 mv^(2)` <br> velocity (v) = `sqrt(2K.E)/m` <br> ` sqrt(2(3.0xx10^(-25)J))/(9.10939xx10^(-31)g)` <br> `sqrt(6.5866xx10^(4)` <br> `v=811.579 ms^(-1)` <br> subsitiuing the value in the expression of `lambda` <br> `lambda =(6.626xx10^(-34) Js)/((9.10939xx10^(-31) kg) (811.5790 ms^(-1)))` <br> `lambda = 8.9625xx 10^(-7) m` <br> Hence, the wavelength of the electron is `8.9625 xx 10^(-7) m `.