Solve Stopping Potential when 250 nm Light Hits Zinc Plate

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SUMMARY

The stopping potential when 250 nm light strikes a zinc plate can be calculated using the equation V = (hc/λ - φ)/q, where h is Planck's constant (6.626 x 10-34 Js), c is the speed of light (3 x 108 m/s), λ is the wavelength (250 nm), φ is the work function of zinc (4.3 eV), and q is the charge of an electron (1.602 x 10-19 C). By substituting these values into the equation, one can determine the stopping potential. The discussion confirms that the problem primarily involves high school physics concepts rather than advanced quantum mechanics.

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Homework Statement



I'm doing some practice problems for an introductory quantum mechanics course and am unsure whether or not I'm solving this problem properly - I need confirmation if I'm doing it right and help if I'm doing it wrong! :)

"What is the stopping potential when 250 nm light strikes a zinc plate?" (Chapter 3, #18 in Modern Physics 2nd ed. by Randy Harris)

Homework Equations



K = E - [itex]\varphi[/itex] (where K is the kinetic energy, E is the energy of the incident light and [itex]\varphi[/itex] is the work function)

This can be written as:

[itex]\frac{mv^2}{2}[/itex] = [itex]\frac{hc}{\lambda}[/itex] - [itex]\varphi[/itex] (where m is the mass of a scattered electron, v is the speed of this electron, h is Planck's constant and c is the speed of light)

[itex]\frac{mv^2}{2}[/itex] = qV (where q is the electron charge and V is the stopping potential)

h = 6.626 x 10^(-34) Js
c = 3 x 10^8 m/s
[itex]\varphi[/itex] = 4.3 eV

The Attempt at a Solution



If my equations above are correct, I can write:

[itex]\frac{mv^2}{2}[/itex] = [itex]\frac{hc}{\lambda}[/itex] - [itex]\varphi[/itex] = qV
[itex]\frac{hc}{\lambda}[/itex] - [itex]\varphi[/itex] = qV
V = [itex]\frac{\frac{hc}{\lambda} - \varphi}{q}[/itex]

I can then simply plug in my values (remembering to either convert h in eV*s or [itex]\varphi[/itex] into J) and this should give me the stopping potential, correct?
 
Last edited:
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Yes, that all looks good. No quantum mechanics in there, though; its all high school physics.
 
Thanks! And you're right, there isn't really any quantum mechanics here; it's just included in part of the course and its part of the introduction leading into the actual quantum mechanics.
 

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