Calculate the wavelength of De Broglie wavelength associated with electrons accelerated through a potential difference of 200 V. (Answer: 0.86 A°)



Given Data:

Potential difference = V = 200 V

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

∴ Charge of electron = e = 1.6 ｘ 10⁻¹⁹ C

∴ 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, to find its value we know that work is done on accelerated an electron through a potential difference of 200 V and this work done is equal to the K.E of the electron. so


K.Emax = W


Where W = q V = e V and K.E = 12 mv²


[ from electrostatic V = wq and here q = e}


12 mv² = e V


or

v² = 2eVm


taking square root we get


v = √2eVm


by putting values


v = √2ｘ1.6ｘ10⁻¹⁹Cｘ200V9.11ｘ10⁻³¹kg


by simplifying we get


v = 8.386ｘ 10⁶ m s⁻¹


Now putting all the corresponding values in equation (1)

λ = 6.626ｘ10⁻³⁴m²kgs⁻¹9.11ｘ10⁻³¹kgｘ8.386ｘ10⁶ms⁻¹




λ = 6.626ｘ10⁻³⁴m²kgs⁻¹76.39646ｘ10⁻²⁵kgms⁻¹

λ = 0.0867 ｘ 10⁻⁹ m

or

λ = 0.867 ｘ 10⁻¹⁰ m

[10⁻¹⁰ m = A°] 

λ = 0.867 A° --------------------Ans.


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