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Calculate the de broglie wavelength associated with an alpha particle accelerated through a potential difference of 200 volt.
I shall work in esu units,as I am familiar with these units.Charge on alpha particles= 2 units of electronic charge =2×4.8×10^-10 esu of charge = 9.6×10^-10 esu of chargeAccelerating voltage= V volts= V/300 esu of potentialEnergy E gained by the alpha particle in being accelerated by the applied potential difference= charge ( in esu) × potential difference in esu = 9.6×10^-10×V/300 = 3.2×10^-12 V ergsNow the kinetic energy E = p²/2m , where p= momentum of the alpha particle, m= mass of the alpha particle =4 units of amu= 4×1.672×10^-24 gm= 6.688×10^-24 gmThe momentum p of the alpha particle = [2 m E]^½=[2×6.688×10^-24 ×3.2×10^-12 V ]^½= [42.803×10^-36 V]^½= 6.542×10^-18 √VTherefore the deBroglie wavelength= h/p= 6.652×10^-27/6.542×10^-18 √V= 1.0167 ×10^-9 cm/√V cm= 0.1016 A°/√V ,now put the value of V = 200 volts
I shall work in esu units,as I am familiar with these units.
Charge on alpha particles= 2 units of electronic charge =2×4.8×10^-10 esu of charge = 9.6×10^-10 esu of charge
Accelerating voltage= V volts= V/300 esu of potential
Energy E gained by the alpha particle in being accelerated by the applied potential difference= charge ( in esu) × potential difference in esu = 9.6×10^-10×V/300 = 3.2×10^-12 V ergs
Now the kinetic energy E = p²/2m , where p= momentum of the alpha particle, m= mass of the alpha particle =4 units of amu= 4×1.672×10^-24 gm= 6.688×10^-24 gm
The momentum p of the alpha particle = [2 m E]^½=[2×6.688×10^-24 ×3.2×10^-12 V ]^½= [42.803×10^-36 V]^½= 6.542×10^-18 √V
Therefore the deBroglie wavelength= h/p= 6.652×10^-27/6.542×10^-18 √V= 1.0167 ×10^-9 cm/√V cm
= 0.1016 A°/√V ,
now put the value of V = 200 volts
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