What is the stoping voltage

[bayan]

Hi everyone.

I came across with a problem in my question book which was

"what is the stoping voltage when a UV ligh with WaveLength 200nm hits a clean gold surface.(5.1v Work Function.)"


I have found the frequency which is $$1.5^15 Hz$$ I know the $$W$$ Therefore I can let $$Ekmax=(hf)-W$$ and $$Ekmax=qV$$

But when I put the equations together they don't make any sense to me.
$$V=\frac{(hf)-W}{q}$$

Can someone help me through this please

 

[dextercioby]

1.You should have posted this in the homework section.

2.Why don't they make any sense?

It looks okay so far.

Daniel.

 

[thepatientmental]

The stopping voltage, also known as the cutoff voltage, is the minimum voltage required to stop the flow of electrons from a metal surface when it is exposed to electromagnetic radiation. In this case, the UV light with a wavelength of 200nm is incident on a clean gold surface with a work function of 5.1v. The frequency of the UV light is 1.5^15 Hz.

To calculate the stopping voltage, we can use the equation V = (hf - W)/q, where V is the stopping voltage, h is Planck's constant, f is the frequency of the incident light, W is the work function, and q is the charge of an electron.

In this case, we know all the values except for V. Substituting the given values, we get V = ((6.626 x 10^-34 J*s)(1.5 x 10^15 Hz) - (5.1 eV)(1.602 x 10^-19 J/eV)) / (1.602 x 10^-19 C)

Simplifying, we get V = 2.49 volts. This means that the stopping voltage for this scenario is 2.49 volts. This is the minimum voltage required to stop the flow of electrons from the gold surface when exposed to UV light with a wavelength of 200nm.

I hope this helps clarify the concept of stopping voltage for you. If you have any further questions, please let me know.