Light consisting of two wavelengths, a = 4.5 x 10 -7 meter and b = 6.4 x 10 -7 m

Question

Light consisting of two wavelengths, a = 4.5 x 10-7 meter and b = 6.4 x 10-7 meter, is incident normally on a barrier with two slits separated by a distance d. The intensity distribution is measured along a plane that is a distance L = 0.83 meter from the slits as shown above. The movable detector contains a photoelectric cell whose position y is measured from the central maximum. The first order maximum for the longer wavelength b occurs at y = 1.9 x 10-2 meter.

a. Determine the slit separation d.  m
b. At what position ya does the first order maximum occur for the shorter wavelength a?  m

In a different experiment, light containing many wavelengths is incident on the slits. It is found that the photosensitive surface in the detector is insensitive to light with wavelengths longer than 6.0 x 10-7 m.

c. Determine the work function of the photosensitive surface.  eV
d. Determine the maximum kinetic energy of electrons ejected from the photosensitive surface when exposed to light of wavelength = 4.5 x 10-7 m.  eV

Explanation / Answer

d sin theta = m lamda

d ( y/L) = m lamda

d = 1 * lamda * L/ y

= 1 * 6.4 * 10^-7 * 0.83/1.9 * 10^-2

=2.79 * 10^-5 m

(b)

y= 1( 4.5 * 10^-7 m)(0.83)/2.79 * 10^-5 m

=1.33 * 10^-2 m

(3)

work fuction = 1242/lambda(in nm)= 1242 eV nm /6.0 * 10^-7 m= 2.07 eV

(4)

K_max = hf- w

   = hc/ lamda - work function

   = 1242 eV nm/4.5 * 10^-7 - 2.07 eV

   =0.69 eV

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