Question

What is the answer to part B?

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 0.160 nm ?

Answer: 58.8

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 6.63×10^?34 J?s for Planck's constant, 3.00×10⁸ m/s for the speed of light in a vacuum, and 1.60×10^?19 C for the charge on an electron. Express your answer using three significant figures.

Answer 1

Part A. Already Solved

Using energy conservation: q*dV = (1/2)*m*V^2

from de-broglie wavelength, = h/(m*v), v = h/(m)

dV = (1/2)*(m/q)*(h/(m))^2 = h^2/(2*q*m*^2)

dV = (6.63*10^-34)^2/(2*1.6*10^-19*9.11*10^-31*(0.160*10^-9)^2)

dV = 58.9 V

Part B.

Given that electrons and x-ray have same energy, So energy of x-ray will be:

E = h*c/

E = 6.63*10^-34*3*10^8/(0.160*10^-9) = 1.243*10^-15 J

Since 1 eV = 1.6*10^-19 J, So

E = (1.243*10^-15 J)*(1 eV/(1.6*10^-19 J)) = (1.243*10^4/1.6)

E = 7768.75 eV

Now Using energy conservation, when electrons are accelerated through potential difference than energy of electron will be:

U = q*dV, So

q*dV = E

dV = Potential difference = E/q

q = charge on electrons = e, So

dV = E/e = 7768.75 eV/e

dV = 7768.75 V

In three significant figures

dV = Potential difference = 7770 V = 7.77*10^3 V

Let me know if you've any query.