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So...1 Ev = 1.6 X 10 J

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So...1 Ev = 1.6 X 10 J An electron ­volt (abbreviation: eV) is a small unit of energy used for elementary charges moving through voltage differences in electric fields. An electron volt is the amount of energy change an elementary charge (like an electron or proton) experience when moved a across a potential difference of 1 volt. Using our new equation, W = Vq we can see how many joules one electron­volt is equivalent to..... W = V q = (1 volt) (1.6 x 10­19 C ) = 1.6 x 10­19 Joules So.... 1 eV = 1.6 x 10­19 J on page 1 of reference table You should be able to convert eV's to Joules and Joules to eV's 1.6 x 10­19 Joule examples: 400 eV = 400 eV x ____________ = 6.4 x 10­17 J 1 eV 1 eV ___________ 7 or... 9.6 x 10­12 Joules converted to eV.... 9.6 x 10­12 J x = 6.0 x 10 eV 1.6 x 10­19 J Examples using the electron volt: 1. An proton is moved across 10 Volts. How much work is done? so, since it takes 1 eV to move a proton across 1 volt, it will take 10 eV to move a proton across 10 Volts 2. An electron is moved across 10 volts. How much work is done? Well, electrons are negative, so negative work will be done, moving it, but it is still just as simple as with the proton. The answer is ­ 10 eV. 3. An electron is moved across 500 Volts of potential difference. How much work is done? so it's ­500 eV. ** please note, this is the same as question 34 on the HW. ­ 500 eV = ­ 500 eV x 1.6 x 10­19 J/eV = = ­ 8.0 x 10­17 J 4. A helium nucleus (+2 elementary charges) is moved across a 700 Volt potential difference. What energy change is there? 2 elem q's x 700 V = 1400 eV oil droplets Yes, I am Robert Millikan 15 14 5 13 12 11 4 OIL 10 9 8 3 7 6 2 5 4 3 1 2 1 89° 0 0 Millikan's Oil Drop + Experiment _ V Summary of Millikan's Oil drop experiment: purpose: to Determine the charge of the electron (the elementary charge!) Synopsis: determine the charge on many oil droplets, then find the largest number that is commonly divisible into all the charges found. That would be the charge of one electron. 1. When oil drops are sprayed out the gain some electrons from the nozzle. 2. some droplets fall between the oppositely charged plated and the voltage can be selectively adjusted to "suspend" some of the drops - in essence balancing the downward pull of gravity with an equal but oppositce upward electrical force. 3. When the voltage is removed, the droplet falls at terminal velocity, which can be measured. The terminal veloicity of the droplet can then be used to determine the drop's weight, thereby determining the electric force. 4. Now knowing the electric force and the stregth of the electric field, Millikan could calculated the Amount of chargeon the droplet in Coulombs. 5. Collecting a variety of charges, he could infer the elctron mass by finding the largest # which could divided evenly into all the data. Easy, right?.
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