Surface of certain metal is first illuminated with light of wavelength lambda_1 = 350 n-m and then b

Q: Why did we assume $v_1 = 2v$ and not $v_2 = 2v_1$?

Because the energy of a photon is inversely proportional to its wavelength ($E = \frac{hc}{\lambda}$). A shorter wavelength ($\lambda_1 = 350\text{ nm}$) means higher incident energy, which results in a higher kinetic energy and thus a higher speed for the ejected electron.

Q: What does 'differ by a factor of 2' mean?

It means the ratio of the two speeds is 2. It could mean $v_1 - v_2 = 2$ or $v_1 / v_2 = 2$. In physics problems involving 'factor', it almost always refers to a multiplicative ratio.

Q: Why did we use $hc = 1240\text{ eV}\cdot\text{nm}$?

This is a standard approximation used in modern physics to quickly convert wavelength in nanometers directly into energy in electron-volts. It saves us from dealing with Planck's constant in Joules and the speed of light in meters per second.

Q: Could we have solved this by dividing the two equations instead of substitution?

Yes, dividing the equations is another valid algebraic approach. You would get a ratio on the left side and $4$ on the right side, which you can then cross-multiply to isolate $\phi$. Both methods lead to the exact same result.

Q: What if the calculated work function was negative?

A negative work function is physically impossible. It would mean the metal emits electrons without any incident light! If you get a negative value, it usually means you swapped the velocities (e.g., assuming $v_2 = 2v_1$).

Question

Surface of certain metal is first illuminated with light of wavelength $\lambda_1 = 350	ext{ n-m}$ and then by light of wavelength $\lambda_2 = 540	ext{ n-m}$. It is found that the maximum speed of the photoelectrons in the two cases differ by a factor of 2. The work function of the metal (in eV) is close to
(energy of photon = $\frac{1240}{\lambda(	ext{in n-m})}	ext{ eV}$)

Options:

(A) 5.6

(B) 2.5

(C) 1.8

(D) 1.4

Sigma Percentile · Accepted Answer

The photoelectric effect is one of the most beautiful phenomena in modern physics, bridging the gap between the wave and particle nature of light. In this problem, we are exploring how different wavelengths of incident light affect the kinetic energy—and consequently, the speed—of ejected electrons.

### Decoding the Problem Statement

We are given a single metal surface, which means the **work function ($\phi$)** remains constant throughout the experiment. The metal is illuminated by two different wavelengths:
1. $\lambda_1 = 350 	ext{ nm}$
2. $\lambda_2 = 540 	ext{ nm}$

The problem states that the maximum speeds of the photoelectrons in the two cases differ by a factor of $2$.

Here is the critical logical deduction: **Which electron is faster?** 
We know that the energy of a photon is inversely proportional to its wavelength ($E = \frac{hc}{\lambda}$). Since $\lambda_1 < \lambda_2$, the photons in the first case pack a much stronger punch. Therefore, the electrons ejected in the first case will have a higher kinetic energy and a higher speed.

This allows us to confidently set our velocities as:
$v_1 = 2v$
$v_2 = v$

### The Master Equation

Einstein's photoelectric equation is our primary tool here. It elegantly states that the energy of the incident photon is split into two parts: overcoming the metal's binding energy (work function) and giving the electron kinetic energy.

$$ \frac{hc}{\lambda} = \phi + \frac{1}{2}mv^2 $$

Let's write this equation for both of our cases.

**For Case 1 ($\lambda_1 = 350 	ext{ nm}$):**
$$ \frac{hc}{350} = \phi + \frac{1}{2}m(2v)^2 $$
$$ \frac{hc}{350} = \phi + 4\left(\frac{1}{2}mv^2ight) $$

**For Case 2 ($\lambda_2 = 540 	ext{ nm}$):**
$$ \frac{hc}{540} = \phi + \frac{1}{2}mv^2 $$

### Algebraic Manipulation

We have a system of two equations. Our goal is to find the work function $\phi$. The easiest way to do this is to eliminate the kinetic energy term ($\frac{1}{2}mv^2$).

From the second equation, we can isolate the kinetic energy:
$$ \frac{1}{2}mv^2 = \frac{hc}{540} - \phi $$

Now, we substitute this expression into our first equation:
$$ \frac{hc}{350} = \phi + 4\left(\frac{hc}{540} - \phiight) $$

Let's expand the bracket and group the $\phi$ terms together:
$$ \frac{hc}{350} = \phi + \frac{4hc}{540} - 4\phi $$
$$ 3\phi = \frac{4hc}{540} - \frac{hc}{350} $$
$$ \phi = \frac{hc}{3} \left( \frac{4}{540} - \frac{1}{350} ight) $$

### The Final Calculation

The problem kindly provides the value of $hc = 1240 	ext{ eV}\cdot	ext{nm}$. Let's plug this in and crunch the numbers.

$$ \phi = \frac{1240}{3} \left( \frac{4 	imes 350 - 540}{540 	imes 350} ight) $$
$$ \phi = \frac{1240}{3} \left( \frac{1400 - 540}{189000} ight) $$
$$ \phi = \frac{1240}{3} \left( \frac{860}{189000} ight) $$
$$ \phi \approx 1.88 	ext{ eV} $$

Looking at our options, the closest value is **$1.8 	ext{ eV}$**.

This problem beautifully demonstrates how a simple ratio in physical observables (like speed) can be traced back through conservation of energy to reveal fundamental properties of materials.

Select Answer:

Visualized Solution

\text{Visualizing the Setup}

\text{Einstein's Photoelectric Equation}

\text{Setting up the Equations}

\text{Eliminating Kinetic Energy}

\text{Solving for Work Function } \phi

\text{Final Calculation}

\text{Conclusion}

The Sigma Insight: Photoelectric Effect

Decoding the Problem Statement

The Master Equation

Algebraic Manipulation

The Final Calculation

Similar Questions

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Select Answer:

Visualized Solution

\text{Visualizing the Setup}

\text{Einstein's Photoelectric Equation}

\text{Setting up the Equations}

\text{Eliminating Kinetic Energy}

\text{Solving for Work Function } \phi

\text{Final Calculation}

\text{Conclusion}

The Sigma Insight: Photoelectric Effect

Decoding the Problem Statement

The Master Equation

Algebraic Manipulation

The Final Calculation

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The work function of a substance is 4.0 eV. The longest wavelength of light that can cause photoelectron emission from this substance is approximately

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