A beam of electromagnetic radiation of intensity 6.4 x 10^-5 W/cm^2 is comprised of wavelength lambd

Q: Why do we need to convert the photon energy from eV to Joules? Can't we just work in eV?

We must convert to Joules because the intensity is given in $\text{W/cm}^2$. A Watt is defined as $1 \text{ Joule per second}$. If you divide a total energy in Joules by a single photon energy in eV, your units will mismatch and your final photon count will be completely wrong.

Q: The problem gives the work function $\phi = 2 \text{ eV}$. Why didn't we use it in any of our calculations?

The work function determines the *kinetic energy* of the emitted electrons ($K_{\text{max}} = E - \phi$) and whether emission happens at all. Since our photon energy ($4 \text{ eV}$) is greater than $2 \text{ eV}$, emission occurs. However, the question only asks for the *number* of electrons ejected, which depends purely on the number of incident photons and the quantum efficiency, not on how fast the electrons are moving once they leave.

Q: What does 'one in $10^3$ photons ejects an electron' actually mean physically?

This represents the 'quantum efficiency' of the photoelectric effect. In reality, when light hits a metal, most photons interact with the lattice and dissipate their energy as heat, or they are reflected. Only a very small fraction of photons successfully transfer their energy to an electron in a way that allows it to overcome the surface barrier and escape.

Q: The intensity is in $\text{W/cm}^2$ and the area is in $\text{cm}^2$. Do I need to convert them to $\text{m}^2$?

No, you don't need to convert to $\text{m}^2$ in this specific case. When you multiply Intensity ($\text{W/cm}^2$) by Area ($\text{cm}^2$), the $\text{cm}^2$ units perfectly cancel out, leaving you directly with Watts (Joules/second).

Q: What would happen if the incident wavelength was $700 \text{ nm}$ instead of $310 \text{ nm}$?

If $\lambda = 700 \text{ nm}$, the photon energy would be $E = \frac{1240}{700} \approx 1.77 \text{ eV}$. Because $1.77 \text{ eV}$ is less than the work function of $2 \text{ eV}$, the photons wouldn't have enough energy to eject any electrons. The number of ejected electrons would be exactly zero, regardless of the intensity!

Question

A beam of electromagnetic radiation of intensity $6.4 	imes 10^{-5} 	ext{ W/cm}^2$ is comprised of wavelength $\lambda = 310 	ext{ nm}$. It falls normally on a metal (work function $\phi = 2 	ext{ eV}$) of surface area of $1 	ext{ cm}^2$. If one in $10^3$ photons ejects an electron, total number of electrons ejected in $1 	ext{ s}$ is $10^x$. ($hc = 1240 	ext{ eVnm}$, $1 	ext{ eV} = 1.6 	imes 10^{-19} 	ext{ J}$), then $x$ is ......... .

Sigma Percentile · Accepted Answer

This problem is a beautiful demonstration of how the macroscopic concept of intensity bridges with the microscopic quantum world of photons and electrons. Let's break down the physics step-by-step.

### Analyzing the Setup

Imagine you are observing a metal plate with a surface area of exactly $1 	ext{ cm}^2$. An electromagnetic radiation beam is striking this plate normally (perpendicularly). We are given two crucial pieces of information about this beam: its intensity $I = 6.4 	imes 10^{-5} 	ext{ W/cm}^2$ and its wavelength $\lambda = 310 	ext{ nm}$.

Our ultimate goal is to find out how many electrons are ejected from this metal plate every single second. To do this, we must first understand the energy carried by the individual "packets" of this radiation—the photons.

### The Energy of a Single Photon

The energy $E$ of a single photon is inversely proportional to its wavelength, given by the famous Planck-Einstein relation:

$E = \frac{hc}{\lambda}$

We are given the convenient constant $hc = 1240 	ext{ eVnm}$. Substituting our wavelength $\lambda = 310 	ext{ nm}$ into the equation:

$E = \frac{1240}{310} = 4 	ext{ eV}$

So, each photon carries $4 	ext{ eV}$ of energy. Notice that the work function of the metal is $\phi = 2 	ext{ eV}$. Since the incident photon energy ($4 	ext{ eV}$) is strictly greater than the work function ($2 	ext{ eV}$), photoelectric emission is physically possible!

However, to relate this to the macroscopic intensity (which is measured in Watts, or Joules per second), we must convert this energy from electron-volts to standard SI units (Joules):

$E = 4 	imes 1.6 	imes 10^{-19} 	ext{ J} = 6.4 	imes 10^{-19} 	ext{ J}$

### Total Energy and Photon Flux

Now, let's shift our focus from a single photon to the entire beam. Intensity is defined as the total energy falling on a unit area per unit time. Therefore, the total power $P$ (energy per second) incident on our specific metal plate is the intensity multiplied by the area:

$P = I 	imes A$

Substituting the given values:

$P = (6.4 	imes 10^{-5} 	ext{ W/cm}^2) 	imes (1 	ext{ cm}^2) = 6.4 	imes 10^{-5} 	ext{ W}$

Since $1 	ext{ W} = 1 	ext{ J/s}$, the total energy striking the plate every second is $6.4 	imes 10^{-5} 	ext{ Joules}$.

If we know the total energy arriving per second, and we know the energy of exactly one photon, we can easily calculate the total number of photons arriving per second ($n_p$) by simply dividing the two:

$n_p = \frac{P}{E} = \frac{6.4 	imes 10^{-5}}{6.4 	imes 10^{-19}}$

$n_p = 10^{14} 	ext{ photons/s}$

### The Quantum Efficiency Catch

Here is where many students make a silly mistake. It is tempting to assume that every single photon that hits the metal will eject an electron. But the physical reality is much messier. Most photons are absorbed as thermal energy or scattered.

The problem explicitly states a quantum efficiency: only one in $10^3$ photons successfully ejects an electron. Therefore, the number of ejected electrons per second ($n_e$) is a tiny fraction of the incident photons:

$n_e = \frac{n_p}{10^3} = \frac{10^{14}}{10^3} = 10^{11} 	ext{ electrons/s}$

### Final Calculation

The problem states that the total number of electrons ejected in $1 	ext{ s}$ is $10^x$. By comparing our result with this format:

$10^{11} = 10^x$

**We can confidently conclude that $x = 11$.**

Enter Numerical Value:

Visualized Solution

Visualizing the Setup

$E = \frac{h c}{λ}$

$E = \frac{1240}{310}$

$E = 6.4 \times 1 0^{- 19} J$

$P = I \times A$

$P = (6.4 \times 1 0^{- 5}) \times 1$

$P = 6.4 \times 1 0^{- 5} J/s$

$n_{p} = \frac{P}{E}$

$n_{p} = \frac{6.4 \times 1 0 ^{- 5}}{6.4 \times 1 0 ^{- 19}}$

$n_{p} = 1 0^{14} photons/s$

$n_{e} = \frac{n _{p}}{1 0 ^{3}}$

$n_{e} = 1 0^{11} electrons/s$

$x = 11$

The Way Forward

The Sigma Insight: Photoelectric Effect

Analyzing the Setup

The Energy of a Single Photon

Total Energy and Photon Flux

The Quantum Efficiency Catch

Final Calculation

Similar Questions

The surface a metal is illuminated with the light of 400 nm. The kinetic energy of the ejected photoelectrons was found to be 1.68 eV. The work function of the metal is ( $h c = 1240 eV-nm$ )

In a photoelectric effect experiment, the threshold wavelength of light is $380 nm$ . If the wavelength of incident light is $260 nm$ , the maximum kinetic energy of emitted electrons will be Given, $E (in eV) = \frac{1237}{λ (in nm)}$

Enter Numerical Value:

Visualized Solution

Visualizing the Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } E=λhc​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } E=3101240​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } E=6.4×10−19 J

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P=I×A

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P=(6.4×10−5)×1

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P=6.4×10−5 J/s

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } np​=EP​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } np​=6.4×10−196.4×10−5​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } np​=1014 photons/s

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ne​=103np​​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ne​=1011 electrons/s

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=11