<html>
<head>
<meta content="text/html; charset=ISO-8859-1"
http-equiv="Content-Type">
</head>
<body bgcolor="#FFFFFF" text="#000000">
The energy of the ejected electron after you exceed the work
function of the metal is given by <br>
<br>
E = h nu - phi<br>
<br>
Some older texts expressed this as<br>
<br>
-E = phi - h nu<br>
<br>
to reiterate the point that the electron would not leave the surface
(with a negative energy) unless the energy of the work function was
exceeded by the incident radiation (resulting in a positive
energy). I must have resorted to that equation in my head and
written just left off the initial negative sign. Sorry about that.<br>
<br>
<br>
On 08/28/2012 05:00 PM, wrote:
<blockquote
cite="mid:CA765D0D95A04D449667AFA14377899C555B5CE043@MERCERMAIL.MercerU.local"
type="cite">
<meta http-equiv="Content-Type" content="text/html;
charset=ISO-8859-1">
<meta name="GENERATOR" content="MSHTML 8.00.6001.19154">
<style id="owaParaStyle">P {
MARGIN-TOP: 0px; MARGIN-BOTTOM: 0px
}
</style>
<div style="direction: ltr;font-family: Tahoma;color:
#000000;font-size: 10pt;">
<p>Dr. Pounds,</p>
<p> </p>
<p>On problem 9.3 from Alberty and Silvey, the book uses
E=h*(nu)-(phi). In class you gave us E=(phi)-h*(nu). Which
of these is correct?</p>
<p> </p>
<p>Thanks,</p>
<p><br>
</p>
</div>
</blockquote>
<br>
<br>
<pre class="moz-signature" cols="72">--
Andrew J. Pounds, Ph.D. (<a class="moz-txt-link-abbreviated" href="mailto:pounds@theochem.mercer.edu">pounds@theochem.mercer.edu</a>)
Associate Professor of Chemistry and Computer Science
Mercer University, Macon, GA 31207 (478) 301-5627
</pre>
</body>
</html>