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 = 21.610¹


by simplifying we get


v = 8.386x 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 8.386x 10⁶ m s⁻¹}




λ = \frac {6.626 x 10⁻³⁴ m² kg s⁻¹}{76.39646 x 10⁻²⁵ kg m s⁻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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