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 = `\frac {h}{λ}`


or


λ = `\frac {h}{P}`


but P = mv , So


λ = `\frac {h}{mv}` ------------(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.E_{max}` = W


Where W = q V = e V and K.E = `\frac {1}{2}` mv²


[ from electrostatic V = `\frac {w}{q}` and here q = e}


`\frac {1}{2}` mv² = e V


or

v² = `\frac {2e V }{m}`


taking square root we get


v = `\sqrt frac {2eV}{m}`


by putting values


v = `\sqrt frac {2 ｘ 1.6 ｘ 10⁻¹⁹ C ｘ 200 V}{9.11 ｘ 10⁻³¹ kg}`


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 ï½˜ 10⁻⁹ m

or

λ = 0.867 ï½˜ 10⁻¹⁰ m

[10⁻¹⁰ m = A°] 

λ = 0.867 A° --------------------Ans.


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