The answer to 10 is 3.96x10^5m/s. The answers to 12 are 1.23x10^x10^-9, 7.07 deg
ID: 2255398 • Letter: T
Question
The answer to 10 is 3.96x10^5m/s. The answers to 12 are 1.23x10^x10^-9, 7.07 degrees, and "yes". I need explanations and work.
The metal barium has a work function of 2.48eV. What is the maximum speed of ejected electrons if it is illuminated with light of wavelength 425nm? Which will eject electron* from a metal with higher kinetic energies: a faint high frequency light source or a bright low frequency light source? Explain. You have a beam of electrons. Assume that each electron has a kinetic energy of leV. a: Determine the DeBroglie wavelength of the electrons. The beam is diffracted through a single slit with alpha = 10 -8 m. Determine the angle of the first diffraction minimum. Is the diffraction noticeable? Explain.Explanation / Answer
10. energy assoicted with 425nm is E = hc/L
E = 6.626*10^-34 *3*10^8/425nm
E = 4.677*10^-19 J
or 2.923 eV
change of enrgy = 2.923-2.48 = 0.443 eV
now 0.5mv^2 = 0.443 *1.6*10^-19
v^2 = 2* 0.443*1.6*10^-19/9.11*10^-31
V^2 = 0.155596032 *10^12
V = 3.946 *10^5 m/s
11, high frequency light will eject eelctrons as E = hf, implues higher the frequency higher the KE and velocity.
12. Debroglie wavelength L = h/sqrt(2meV)
L = 6.626*10^-34*/sqrt(2*9.11*10-^-31*1*1.6*10^-19)
L = 1.22 *10^-9 = 1.22 nm
b. d sin theta =mL
sin theta = 1* 1.22 nm/10^-8
theta = 7 degs
yes since diffracted angle is very small , aptternon the screen will be be very closer to the central spot, so that a clear pattern can be seen,
on the other hand if angle is very large, that corresponds to large fringe width and pattern faints