Number of electrons emitted photoelectrically

[PhysicsTest]

Homework Statement

A certain photo surface has a spectral sensitivity of 6mA/W of incident radiation of wavelength 2,537 A. How many electrons will be emitted photoelectrically by a pulse of radiation consisting of 10,000 photons of this wavelength?

Relevant Equations

$\lambda=\frac{1.24} {E_G}$

Spectral response : $\lambda=\frac{1.24} {E_G}$ ->eq(1)
Given, $\lambda = 2537 * 10^{-10} m = 0.2537\mu m$
$E_G = \frac{1.24} {\lambda} = \frac{1.24} {0.2537} = 4.88 eV$

So, i can assume if 10,000 photons illuminate then 10000 electrons will be emitted. But i am mainly confused with the value of 6mA/W. I think need to convert it into eV to know the cutoff wavelength at which the electron can be excited?

 

[BvU]

So, i can assume if 10,000 photons illuminate then 10000 electrons will be emitted.

Based on what, precisely ?

 

[PhysicsTest]

From the text it says if a photon of high energy illuminates then an electron-hole pair can be created. Based on that if the 10,000 photons have sufficient energy then 10,000 electrons can be generated, that is the idea. Am i correct?

 

[Steve4Physics]

From the text it says if a photon of high energy illuminates then an electron-hole pair can be created. Based on that if the 10,000 photons have sufficient energy then 10,000 electrons can be generated, that is the idea. Am i correct?

View attachment 276153
So, i can assume if 10,000 photons illuminate then 10000 electrons will be emitted.

[Minor edit.]

No, you can't assume this. If this were true, the answer to the original question would simply be 10,000!

Your attachment says ‘can be created’ and uses the word ‘may’, indicating the process is not 100% efficient. For 10,000 photons, *less than* 10,000 electrons will be released.

Note that the original question is basically about knocking out-electrons from the surface of a conductor. Your attachment seems to be about exciting electrons inside a semiconductor, creating electron-hole pairs. You are mixing up two different (though related) processes - which is a potential cause of confusion.

Here are 3 questions for you to consider:
1) What is the energy in the 10000 photons?
2) Suppose the pulse of radiation lasts t seconds, what power (in watts) is delivered by the pulse? (The answer will be an expression which includes ‘t’.)
3) Using the given sensitivity (6mA/W) what current is produced during the t seconds? (The answer will be an expression which includes ‘t’.)

If you can answer those, you are well on the way to completing the question.

 

[Charles Link]

They give you the spectral sensitivity in amperes/watt, and that is the same as coulombs/joule. The two ways of stating the units are identical. (These same units are also useful in working with photodiodes).

 

[PhysicsTest]

Here are 3 questions for you to consider:
1) What is the energy in the 10000 photons?

If i use the eq(1) in post 1. Then the energy of let $n$ photons is $E_G= \frac{1.24n} {\lambda}$ eV ->eq2

2) Suppose the pulse of radiation lasts t seconds, what power (in watts) is delivered by the pulse? (The answer will be an expression which includes ‘t’.)

$E=P*t = P=\frac{E} {t}$ ->eq3.1
$P = \frac{1.24n} {\lambda t}$ ->eq3.2

3) Using the given sensitivity (6mA/W) what current is produced during the t seconds? (The answer will be an expression which includes ‘t’.)

$I^2 R = \frac{1.24n} {\lambda t}$ -> eq4.1
R is resistance
$I = \sqrt{(\frac{1.24n} {\lambda Rt})}$ ->eq4.2
If $s$ is the sensitivity
$s = \frac{I} {P}$ ->eq5.1
I am not sure if i am proceeding correctly?

 

[BvU]

3.2) I agree

4.1) you bring in $R$ which is not present in the problem. Instead you want to take the power in Watts from 3.2 and multiply with 6 mA/Watt

 

[PhysicsTest]

Ok, the current is then if $s$ is the sensitivity
$I = P*s$ using eq3.2
$I = \frac{1.24ns}{\lambda t}$ ->eq5.2
If i use the definition of current as $I =\frac{Q} {t}$
$Q = \frac{1.24ns} {\lambda}$ ->eq5.3
The number of electrons is
$\frac{Q} {q} = \frac{1.24ns} {\lambda q}$ ->eq5.4

 

[BvU]

$P$ is not in Watts yet !

 

[Steve4Physics]

Ok, the current is then if $s$ is the sensitivity
$I = P*s$ using eq3.2
$I = \frac{1.24ns}{\lambda t}$ ->eq5.2
If i use the definition of current as $I =\frac{Q} {t}$
$Q = \frac{1.24ns} {\lambda}$ ->eq5.3
The number of electrons is
$\frac{Q} {q} = \frac{1.24ns} {\lambda q}$ ->eq5.4

Just to expand on what @BvU said (Post #9)...

Your overall logic is correct. But if you are going to put values into a formula, units must match

Sensitivity is given in SI units. In the SI system a watt means 1 joule/second. So you can't use the value of photon energy in unts of eV, it must be in joules.

 

[PhysicsTest]

Yes, I always get confused with the units, the equation 3.2 now modified into
$P = \frac{1.24nq} {\lambda t}$ Watts
The number of electrons is then
$\frac{1.24ns} {\lambda}$

 

[BvU]

I wholeheartedly second @Steve4Physics :

You have a tendency to use expressions without paying attention to the units, thereby creating an abundance of error opportunities. Striking example is the single relevant equation in post #1: $\lambda=\frac{1.24} {E_G}$ where you not only have to remember the expression, but also that $\lambda$ has to be expressed in microns. This is all very well if you are a professional spectroscopist (although some do everything in cm^-1), but not for a general physicist or student.

I urgently recommend you to use SI units unless explicitly mentioned otherwise, and to remember the basic expressions in the simplest possible form in SI units.

In particular, for this exercise:
$$\begin{align}E &= h\nu\\c &= \lambda \nu \end{align}$$
so that $\ \$10 k photons is $\ \ 10^4 \ \displaystyle {hc\over \lambda}\$ J

Multiply with 6 mA/W = 6 mC/J and you have the number of Coulombs per 10k photons.

Divide by e and there's your answer. So simple !

Yes, I always get confused with the units, the equation 3.2 now modified into
$P = \frac{1.24nq} {\lambda t}$ Watts
The number of electrons is then
$\frac{1.24ns} {\lambda}$

And how do you check the dimensions in this answer ?