A metal surface is illuminated by light of two different wavelengths 248 nm and 310 nm. The maximum

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

It is a convenient approximation for the product of Planck's constant ($h$) and the speed of light ($c$). Using this value allows us to directly calculate photon energy in electron-volts (eV) when the wavelength is given in nanometers (nm), saving us from tedious unit conversions.

Q: Why is the kinetic energy ratio the square of the speed ratio?

The formula for kinetic energy is $K = \frac{1}{2}mu^2$. Since the mass of the electron ($m$) and the factor $\frac{1}{2}$ are constant, the kinetic energy is directly proportional to the square of the speed ($u^2$). Therefore, $\frac{K_1}{K_2} = \frac{u_1^2}{u_2^2}$.

Q: What does the work function represent physically?

The work function ($W$) is the minimum amount of energy required to remove an electron from the surface of a specific metal. It is a characteristic property of the material.

Q: Why are the speeds referred to as 'maximum speeds'?

Electrons inside a metal are bound by varying amounts of energy. The 'maximum speed' corresponds to the electrons that were least tightly bound (requiring exactly the work function energy to escape). Electrons that were more tightly bound will emerge with less kinetic energy and therefore lower speeds.

Q: Can the work function be negative?

No, the work function is always a positive value. It represents the energy barrier that must be overcome to free an electron. A negative work function would imply that electrons spontaneously leave the metal without any energy input, which violates the conservation of energy.

Question

A metal surface is illuminated by light of two different wavelengths $248	ext{ nm}$ and $310	ext{ nm}$. The maximum speeds of the photoelectrons corresponding to these wavelengths are $u_1$ and $u_2$, respectively. If the ratio $u_1 : u_2 = 2 : 1$ and $hc = 1240	ext{ e V nm}$, the work function of the metal is nearly

Options:

(A) 3.7 eV

(B) 3.2 eV

(C) 2.8 eV

(D) 2.5 eV

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine a metal surface being struck by two different beams of light. One has a wavelength of $248	ext{ nm}$, and the other has a wavelength of $310	ext{ nm}$. Each beam ejects photoelectrons, but with different maximum speeds. We are given that the ratio of these maximum speeds is $u_1 : u_2 = 2 : 1$. Our goal is to find the work function $W$ of the metal.

### The Master Equation

To solve this, we need **Einstein's photoelectric equation**. The maximum kinetic energy of the emitted electrons, $K_{	ext{max}}$, is equal to the energy of the incident photon, $E$, minus the work function of the metal, $W$.

$$K_{	ext{max}} = E - W$$

Where the energy of the incident photon is given by $E = \frac{hc}{\lambda}$.

### Calculating Photon Energies

Let's calculate the energy of the first photon, $E_1$. Using the given value of $hc$ as $1240	ext{ eV nm}$, we divide it by the first wavelength, $248	ext{ nm}$.

$$E_1 = \frac{1240	ext{ eV nm}}{248	ext{ nm}} = 5	ext{ eV}$$

Similarly, for the second photon, $E_2$, we divide $1240$ by $310	ext{ nm}$.

$$E_2 = \frac{1240	ext{ eV nm}}{310	ext{ nm}} = 4	ext{ eV}$$

### Relating Kinetic Energy to Speed

Now, kinetic energy is proportional to the square of the speed ($K = \frac{1}{2}mu^2$). We are given that the ratio of the maximum speeds, $u_1$ to $u_2$, is $2 : 1$. Therefore, the ratio of their maximum kinetic energies, $K_1$ to $K_2$, will be the square of this ratio.

$$\frac{K_1}{K_2} = \frac{u_1^2}{u_2^2} = \left(\frac{2}{1}ight)^2 = 4$$

### Final Calculation

Let's substitute our energy expressions into this ratio. $K_1$ is $5 - W$, and $K_2$ is $4 - W$. Setting their ratio equal to $4$ gives us a simple algebraic equation to solve for the work function.

$$\frac{5 - W}{4 - W} = 4$$

Cross-multiplying, we get:

$$5 - W = 4(4 - W)$$

$$5 - W = 16 - 4W$$

Rearranging the terms, we find:

$$3W = 11$$

$$W = \frac{11}{3} \approx 3.67	ext{ eV}$$

Rounding off to one decimal place, the work function of the metal is nearly **$3.7	ext{ eV}$**. This matches option (a) perfectly.

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Visualized Solution

$Photoelectric Effect Setup$

$Einstein \overset{s}{ˊ} Photoelectric Equation$

$Energy of First Photon$

$Energy of Second Photon$

$Kinetic Energy Ratio$

$Substituting Energies$

$Solving for Work Function$

$Final Answer$

The Sigma Insight: Photoelectric Effect

Analyzing the Setup

The Master Equation

Calculating Photon Energies

Relating Kinetic Energy to Speed

Final Calculation

Similar Questions

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Photoelectric Effect Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Einsteinsˊ Photoelectric Equation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Energy of First Photon

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Energy of Second Photon

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Kinetic Energy Ratio

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Solving for Work Function

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