What is the de Broglie wavelength of an electron whose kinetic energy is 120 eV?  (Answer: 1.12 ｘ10⁻¹⁰ m)



Given Data:

Kinetic Energy = K.E = 120  eV = 120ｘ 1.6 ｘ 10⁻¹⁹ J = 192 ｘ 10⁻¹⁹ J

∴ Mass of electron = m = 9.11 ｘ 10⁻³¹ kg

∴ Plank's Constant = h =  6.626 ｘ 10⁻³⁴ m² kg s⁻¹

To Find:

De Broglie Wavelength = λ =  ? 




Solution:


We know that momentum P is given by

P = hλ


or


λ = hP


but P = mv , So


λ = hmv ------------(1)


But the value of speed v is unknown so, to find its value we have the formula of K.E


12 mv² = K.E

or

v² = 2K.Em

or

v = √2K.Em


by putting values


v = 2ｘ192ｘ10⁻¹


v = \sqrt frac {384 ｘ 10⁻¹⁹ J}{9.11 ｘ 10⁻³¹ kg}

v = \sqrt {42.151 ｘ 10¹² m² s⁻²}


v = 6.492ｘ 10⁶ m s⁻¹


Now putting all the corresponding values in equation (1)

λ = \frac {6.626 ｘ 10⁻³⁴ m² kg s⁻¹}{9.11 ｘ 10⁻³¹ kg ｘ 6.492ｘ 10⁶ m s⁻¹}


λ = \frac {6.626 ｘ 10⁻³⁴ m² kg s⁻¹}{59.142 ｘ 10⁻²⁵ kg m s⁻¹}

λ = 0.112 ｘ 10⁻⁹ m

or

λ = 1.12 ｘ 10⁻¹⁰ m -------------Ans.

[10⁻¹⁰ m = A°] 

λ = 0.867 A° --------------------Ans.


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