How to answer this photoelectric effect question?

Jan 2020
16
0
Mumbai(Bombay), Maharashtra State, India.
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My attempt:- 1)At 0.68e15 Hz frequency of light,Metal A gives 7.2eV Kinetic Energy(KE).

2)At 1e15 Hz frequency of light, Metal B, gives 6eV Kinetic Energy.

3) At 1.1e15 Hz frequency of light, Metal C gives 5.5eV Kinetic Energy.

4)At1.2e15 Hz frequency of light, Metal D gives 5eV Kinetic energy.

Now $KE_{max}=(\frac{hc}{\lambda}-\phi)$ where $\lambda $=wavelength of light. $\phi$= work function of metal and c= speed of light.

Now $7.2eV=(4.14e-15eVs \times 0.68e15 Hz -\phi) $for Metal A. The wavelength of light is 441 nm. Likewise we can compute wavelength of light incident on Metal B, C and D.

But we don't know the work functions of Metal A,B, C and D.

So, how to answer this question?
 
Last edited:
Jan 2020
16
0
Mumbai(Bombay), Maharashtra State, India.
After performing some computational work, I computed answer to this question which is B.
 

topsquark

Forum Staff
Jul 2013
73
6
Any place where Alyson Hannigan can find me
It is more common to express the equation as \(\displaystyle KE_{max} = hf - \phi\). It makes this a little easier to work with in this case, though there is no real difference.

The work function is the "y" intercept of this equation. So what is the equation for each of the lines?

For example, for metal A, we have two points on the line (I'll call this \(\displaystyle (f, KE_{max} )\) as an ordered pair.)
(0.64, 0) and (2.4, 7.2) (Of course we can use any two points on the line.) So the equation of the line is \(\displaystyle KE_{max} = hf - 2.723 \text{eV}\), so the work function is \(\displaystyle \phi = 2.726 \text{eV}\).

Can you get the other three?

-Dan

Addendum: I see you did some work while I was posting. I get metal B as well, but when I calculated the work function of metal B I get 3.93 eV and when I use the data for the unknown metal I get 4.14 eV, so B is the closest answer. I suppose it could be a rounding error.
 
Last edited:
Jan 2020
16
0
Mumbai(Bombay), Maharashtra State, India.
It is more common to express the equation as \(\displaystyle KE_{max} = hf - \phi\). It makes this a little easier to work with in this case, though there is no real difference.

The work function is the "y" intercept of this equation. So what is the equation for each of the lines?

For example, for metal A, we have two points on the line (I'll call this \(\displaystyle (f, KE_{max} )\) as an ordered pair.)
(0.64, 0) and (2.4, 7.2) (Of course we can use any two points on the line.) So the equation of the line is \(\displaystyle KE_{max} = hf - 2.723 \text{eV}\), so the work function is \(\displaystyle \phi = 2.726 \text{eV}\).

Can you get the other three?

-Dan

Addendum: I see you did some work while I was posting. I get metal B as well, but when I calculated the work function of metal B I get 3.93 eV and when I use the data for the unknown metal I get 4.14 eV, so B is the closest answer. I suppose it could be a rounding error.
Hello,

I computed the answer as follows:-$KE_{max}+\phi=hf\tag{1}$, where hf is energy of light,h=Planck's constant.

Now $KE_{max},\lambda$ are 2.5eV and 1.87e-7 m respectively. We computed work function$\phi$ of metal B is 4.14 eV. Plugging in these values in (1) we get

$2.5 eV+4.14eV=\frac{6.626e-34 J*s*299792458m/s}{1.87e-7m} \Rightarrow 6.63eV=6.63eV$ rounded off to significant figures.

So, we can conclude that Metal B is the unknown metal we are looking for.