Question

A sample is exposed to green light with the energy of 2.4 eV and electron beam of 100 KeV

energy.

a) Calculate the wave length of the photons and electrons in this experiment. Which signal

has a higher wave length? Explain. (10 points)

b) Assuming a convergence angle of 30 degrees for the aperture used in an imaging

experiment for these waves, calculate the resolution for the photon and electron beam in

this experiment (assuming there is no lens aberrations and µ=1). Which wave results in

higher resolution? Explain. (10 points)

Hint: for electrons:

(m: mass of electron, 9.11x10-31 kg, h = 6.63x10-34 m2

kg/s2

)

Answer 1

a)

i) wavelength of photon

Energy of photon ,E = hv

where,

v = frequency = c/wavelength

c = speed of light , 2.99×10^8m/s

h = planck constant , 4.136×10^-15 eV.s

therefore,

2.24ev = 4.136 × 10^-15 eV.s × 2.99×10^8(m/s)/wavelength

wavelength = 4.136 × 10^-15 eV.s × 2.99×10^8(m/s)/2.24eV

= 5.5208 ×10^-7m

= 552.08× 10^-9 m

= 552.08nm

ii) wavelength of electron

De Broglie wavelength , = h/mv

where,

h = planck constant , 6.626 × 10^-34 kg m^2 /s

m = mass of electron , 9.11 × 10^-31kg

v = speed of electron

Kinetic energy of electron , E = 1/2 m v^2 = 100000eV=1.602 × 10^-14J = 1.602 × 10^-14 kg ( m^2/s^2)

1.602×10^-14(kgm^2/s^2)= 1/2 × 9.11×10^-31kg × v^2

v^2 = 2 × 1.602 × 10^-14(kg m^2/s^2)/9.11×10^-31kg

v^2 = 3.517 × 10^16 m^2/s^2

v = 1.875 × 10^8m/s

Therefore,

wavelength

= 6.626 ×10^-34 kg m^2/s/9.11×10^-31kg × 1.875×10^8m/s

= 3.879× 10^ -11m

Therefore,

wavelength of electron = 3.879×10^-11m

iii) wavelength of photons is the higher wave lengths