Question

Question 3: Electron microscopy (7.5 pts)

We use electron microscopes to image nanoscale objects, ranging from cells to gold nanoparticles. In this problem we will use the energy unit of electron volts (or eV). 1eV=1.602*10-19 J. To make thingseasier, you can use Planck’s constant in eV.

? = ?. ???? × ??−???? ?

3a) To generate the electrons used in electron microscopy, we make use of the photoelectric effect. A 150 nm UV photon hits a copper plate generating an electron. First, find the energy of the 200 nm photon in eV.

3b) For this question, you will need the use the equation for the photoelectric effect.

??h???? = ?????????? + ?

Determine the kinetic energy (in eV) of the electron emitted from the copper. The work function of the copper is (? = 4.7 ??).

3c) Convert the kinetic energy of the electron from the electron volts back to joules.

3d) What is the speed of the electron emitted from the metal in m/s?

3e) In this question you will need to use DeBroglie’s Equation.

?=h??

What is the wavelength of the electron emitted from the metal in part 3d?

3f) Viruses have a size of approximately 10 nm. Can we observe (or resolve) viruses with our electron beam? How might we decrease the wavelength of the electron beam to study smaller particles?

Answer 1

We use electron microscopes to image nanoscale objects, ranging from cells to gold nanoparticles. In this problem we will use the energy unit of electron volts (or eV). 1eV=1.602*10-19 J. To make thingseasier, you can use Planck’s constant in eV.

? = ?. ???? × ??−???? ?

3a) To generate the electrons used in electron microscopy, we make use of the photoelectric effect. A 150 nm UV photon hits a copper plate generating an electron. First, find the energy of the 200 nm photon in eV.

Ans): energy of the 200 nm photon in eV

Formula :

Energy (E)= c*h = speed constant*Planck constant in ev   

                      λ                            Wavelength in m

                            = (3*10˄8)*( 4.1357*10˄-15)

                                                 200*10˄-9

Energy of the 200 nm photon = 6.2039 ev

3b) For this question, you will need the use the equation for the photoelectric effect.

??h???? = ?????????? + ?

Determine the kinetic energy (in eV) of the electron emitted from the copper. The work function of the copper is (? = 4.7 ??).

Ans) : Kinetic energy (in eV) of the electron emitted

Formula : E_k = hf- ?   (equation for the photoelectric effect.)

                

            Here, E_k – Kinetic energy.

h – Planck`s Constant in ev; f – Frequency in Hertz ; ? – Work Function

Given: work function of the copper is (? = 4.7 ??).

Firstly need to calculate Frequency

So, f= c = Speed of Light = 3*10˄8          = 2*10˄15 Hz.

           λ     Wavelength          150*10˄-9

Now, E_k = {( 4.1357*10˄-15) * (2*10˄15)} – 4.7

             E_k = 3.57ev

3c) Convert the kinetic energy of the electron from the electron volts back to joules.

   Ans: As we Know,   1eV=1.602*10-19 J

                So 3.57ev = 5.72*10-19 J

3d) What is the speed of the electron emitted from the metal in m/s?

Ans: K.E. = 1/2mv²

              

v = √ 2KE / m

here, m = 9.11* 10-31kg ; KE = 5.72*10-19 J

i.e. v   = √ (2* 5.72*10-19)

                       9.11*10-31

                  = √ 1.2558*10˄12

                  = 1.12*10˄6 m/s

3e) In this question you will need to use DeBroglie’s Equation.

?=h??

What is the wavelength of the electron emitted from the metal in part 3d?

Ans: DeBroglie’s Equation : λ = h/mv

Here: h=6.626*10-34J; m= 9.11*10-31; v= 1.12*10˄6

    λ = h/mv

     =                         6.626*10-34   = 0.649*10-9 m

(9.11*10-31)*( 1.12*10˄6)

    Wavelength of the electron emitted from the metal in part 3d = 0.649 nm

3f) Viruses have a size of approximately 10 nm. Can we observe (or resolve) viruses with our electron beam? How might we decrease the wavelength of the electron beam to study smaller particles?

Ans: yes we can observe . but we nned a higher resolution Microscope

Electron Microscope has a range of max 200 nm

We might need higher energy emitting plates by which we can detect the different virus sizes.