1. An electron is accelerated through a potential difference of 3477 Volts in an
ID: 1535727 • Letter: 1
Question
1. An electron is accelerated through a potential difference of 3477 Volts in an electron microscope. Find the electron's velocity and use it to determine the De Broglie wavelength of the electrons in the microscope's beam in meters.
a. A metal surface illuminated by a light source with a variable frequenct exhibits the photoelectric effect. It is found that electrons only come off the metal surface when the wavelength of light is less than 264nm. What is the work function for this metal in electron volts?
b. Suppose than an electron in hydrogen has been excited to the n=5 orbit. What wavelength of light would be emitted if the electron transitioned from the n-5 to n=2 orbits? Give answers in nm
c. A positive point charge of 3.3x10^-7 C at the point (x,y)=(-1.1/2m, 0 m) is 1.1m away from a negative point charge of 3.3x10^-7 C at (x,y)=(+1.1/2m, 0 m) on the x-axis. Find the electric field (size and direction) at a point on the y-axis (x,y)=(0m, 3.5 m) due to the two charges.
Explanation / Answer
1)
Workdone on electron = gain in kinetic energy
q*delta_V = (1/2)*m*v^2
v = sqrt(2*q*delta_V/m)
= sqrt(2*1.6*10^-19*3477/(9.11*10^-31))
= 3.49*10^7 m/s
Broglie wavelength, lamda = h/(m*v)
= 6.626*10^-34/(9.11*10^-31*3.49*10^7)
= 2.08*10^-11 m
a) Workd function, Wo = h*c/lamda
= 6.626*10^-34*3*10^8/(264*10^-9)
= 7.53*10^-19 J
= 7.53*10^-19/(1.6*10^-19)
= 4.71 eV
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