Part A:- Through what potential difference V must electrons be accelerated (from

Question

Part A:-  Through what potential difference V must electrons be accelerated (from rest) so that they will have the same wavelength as an x-ray of wavelength 0.170 nm ? Use 6.63×1034 Js for Planck's constant, 9.11×1031 kg for the mass of an electron, and 1.60×1019 C for the charge on an electron. Express your answer using three significant figures.

Part B:- Through what potential difference V must electrons be accelerated so they will have the same energy as the x-ray in Part A? Use 6.63×1034 Js for Planck's constant, 3.00×108 m/s for the speed of light in a vacuum, and 1.60×1019 C for the charge on an electron. Express your answer using three significant figures.

Explanation / Answer

Part A

momentum is p = h/lamda

but p = sqrt(2*K*m)

kinetic energy K = q*delta V = 1.6*10^-19*delta V

then

p = sqrt(2*K*m)

h/lamda = sqrt(2*1.6*10^-19*delta V*9.11*10^-31)

(6.63*10^-34)/(0.17*10^-9) = sqrt(2*1.6*10^-19*delta V*9.11*10^-31)

delta V = 52.2 V

Part B)

Energy of x-ray is E = h*c/lamda = (6.63*10^-34*3*10^8)/(0.17*10^-9) = 1.17*10^-15 J

E = q*delta V

1.17*10^-15 = 1.6*10^-19*deltaV

delta V = 7312.5 V

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