The answer to 10 is 3.96x10^5m/s. The answers to 12 are 1.23x10^x10^-9, 7.07 deg

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

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