Divided Photo
Divided Photo
Photo-electric effect of Zinc?
What wavelength of light should fall on the zinc (4.31 eV work function) if you emitting electrons with a maximum speed of 1.00E 6 m / s? Given the surprising speed in KE =. 5 (mass) (speed) which is equivalent square 0.5 (9.91e-31), (1e12) = 4.955e-19 J. Which becomes divided by 1.6E -19 eV for KE is our 3.097eV. Getting to E = KE + W. KE + W is 3.097 + 4.31 is 7.41eV. E = hc / Lambda. (H) (c) is 1240eV to divide 1240 by lambda is equal to our KE + W. 1240 divided by 7.41 leads 167nm our response. This is wrong and not know why. Any help would be greatly appreciated. THANK YOU-I 'll give 10 points "I see what you are saying, we are calculating the threshold wavelength for the photoelectric effect. What everything we do is to remove the electron so that our W = E. But with this, I give the wrong answer of 287.7
Using the formula E = hv, as he did. Not Ignore the electron velocity, as it is included in the work function. So use the value of 4.31 eV only.
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