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Chapter 17: Problem 17

A potassium surface is placed \(75 \mathrm{~cm}\) away from a \(100 \mathrm{~W}\) bulb. It is found that the energy radiated by the bulb is \(5 \%\) of the input power. Consider each K-atom as a circular disc of diameter \(1 \AA\). Determine the time required for each atom to absorb an amount of energy equal to its work function of \(2.0 \mathrm{eV}\) considering wave nature of light. n

Short Answer

Expert verified
The time required is approximately 577 seconds.

Step by step solution

01

Determine the Power Radiated by the Bulb

The input power of the bulb is given as 100 W. Since the bulb radiates only 5% of this power, the radiated power \( P_{r} \) can be calculated as \( P_{r} = 0.05 \times 100 \text{ W} = 5 \text{ W} \).
02

Calculate the Intensity of Light at the Potassium Surface

The intensity \( I \) of light at the surface of the potassium is given by the expression \( I = \frac{P_{r}}{A} \), where \( A \) is the area through which the power radiates. Since the bulb is spherical light source, \( A = 4\pi r^2 \), where r is the distance from the bulb. Substituting the values, \( A = 4 \pi (0.75 \text{ m})^2 = 7.06858 \text{ m}^2 \). Then the intensity \( I \) is given by \( I = \frac{5 \text{ W}}{7.06858 \text{ m}^2} \approx 0.707 \text{ W/m}^2 \).
03

Calculate the Area of a K Atom

The area \( A_k \) of a potassium atom, modeled as a circular disc, can be calculated using the formula for the area of a circle, \( A = \pi r^2 \). With the diameter given as \( 1 \text{ 脜} = 1 \times 10^{-10} \text{ m} \), the radius \( r \) is \( 0.5 \times 10^{-10} \text{ m} \). Hence, \( A_k = \pi (0.5 \times 10^{-10})^2 \approx 7.854 \times 10^{-21} \text{ m}^2 \).
04

Compute the Energy Absorbed per Second by One K Atom

The energy absorbed by a single potassium atom per second is the product of the intensity \( I \) and the area \( A_k \) of the atom. Therefore, the power absorbed by one potassium atom \( P_a = I \times A_k = 0.707 \text{ W/m}^2 \times 7.854 \times 10^{-21} \text{ m}^2 \approx 5.55 \times 10^{-21} \text{ W} \).
05

Calculate the Time Required to Absorb the Energy Equivalent to the Work Function

The work function is given as \( 2.0 \text{ eV} \). Converting this energy to joules, \( 1 \text{ eV} = 1.602 \times 10^{-19} \text{ J} \), so \( 2.0 \text{ eV} = 3.204 \times 10^{-19} \text{ J} \). The time \( t \) required to absorb this energy is given by \( t = \frac{\text{Energy}}{\text{Power}} = \frac{3.204 \times 10^{-19} \text{ J}}{5.55 \times 10^{-21} \text{ W}} \approx 577 \text{ seconds} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Work Function
The work function is one of the essential concepts when discussing the photoelectric effect. It represents the minimum energy needed to remove an electron from the surface of a material. For potassium, the work function is given as 2.0 eV in the exercise. This means that each potassium atom requires this amount of energy to eject an electron.

Let's break this down further:
  • An electron is bound to an atom by energy. To release this electron, that bond must be overcome, which is quantified by the work function.
  • In practical terms, this energy is typically provided by incident light in the form of photons.
  • If the absorbed photon has energy equal to or greater than the work function, the electron is ejected.
Understanding the work function allows us to predict if and how quickly electrons can be released from a surface when exposed to light energy.
Intensity of Light
In this context, intensity refers to the power of light passing through a given area. It is a crucial element in understanding how light affects materials like potassium.

Consider a light source spreading its energy across a spherical surface. The intensity of light at any distance can be determined using the formula:
  • \( I = \frac{P_r}{A} \)
  • This refers to power radiated divided by the area of the sphere at that distance.


In our exercise, the light source is a bulb, and the potassium surface is 75 cm away. We've calculated that the intensity at this distance is roughly 0.707 W/m虏.
This measurement is fundamental for understanding the power absorbed by individual atoms.
Energy Absorption
Energy absorption is the process by which a material, like potassium, takes in energy, here from light. When potassium atoms absorb light energy, under the right conditions, they have enough energy to eject electrons.

Let's detail how this happens:
  • The energy absorbed by each potassium atom is the product of the light's intensity and the atom's surface area. Practically, this is calculated as:
    \( P_a = I \times A_k \)
  • With the given values, each potassium atom absorbs approximately \(5.55 \times 10^{-21}\text{ W} \)
Understanding energy absorption is crucial because it determines the feasibility and time required for an atom to release an electron through the photoelectric effect.
Power Radiation
Power radiation relates to the energy transmitted by a light source in the form of electromagnetic waves. In the given problem, a bulb has an input power of 100 W. However, not all of it is radiated as useful energy. Only 5% is effectively radiated, calculated as:

\( P_{r} = 0.05 \times 100 W = 5 W \)

This is the energy spread across the area that reaches the potassium atoms. Understanding power radiation helps set the stage for how much electric force is available from the bulb.
  • This value takes into account inefficiencies like energy lost in the form of heat.
  • The effective power (5 W) dictates both the intensity of the light and eventually factors into how quickly energy can be absorbed by each atom on the surface.
By grasping power radiation, students can better understand the limits and behavior of light impacting atomic interactions and electron ejection.

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Most popular questions from this chapter

A photon scatters in the backward direction \(\left(\phi=180^{\circ}\right)\) from a free proton that is initially at rest. What must the wavelength of the incident photon be if it is to undergo a \(10.0 \%\) change in wavelength as a result of the scattering?

A photon with wavelength \(0.1100 \mathrm{~nm}\) collides with a free electron that is initially at rest. After the collision the wavelength is \(0.1132 \mathrm{~nm}\). (a) What is the kinetic energy of the electron after the collision? What is its speed? (b) If the electron is suddenly stopped (for example, in a solid target), all of its kinetic energy is used to create a photon. What is the wavelength of this photon?

The cathode-ray tubes that generated the picture in early color televisions were sources of \(x\) rays. If the acceleration voltage in a television tube is \(15.0 \mathrm{kV}\), what are the shortest-wavelength \(x\) rays produced by the television? (Modern televisions contain shielding to stop these \(x\) rays.)

A 2.50-W beam of light of wavelength \(124 \mathrm{~nm}\) falls on a metal surface. You observe that the maximum kinetic energy of the ejected electrons is \(4.16 \mathrm{eV}\). Assume that each photon in the beam ejects a photoelectron. (a) What is the work function (in electron volts) of this metal? (b) How many photoelectrons are ejected each second from this metal? (c) If the power of the light beam, but not its wavelength, were reduced by half, what would be the answer to part (b)? (d) If the wavelength of the beam, but not its power, were reduced by half, what would be the answer to part (b)?

(a) A proton is moving at a speed much slower than the speed of light. It has kinetic energy \(K_{1}\) and momentum \(p_{1}\). If the momentum of the proton is doubled, so \(p_{2}=2 p_{1}\), how is its new kinetic energy \(K_{2}\) related to \(K_{1} ?\) (b) A photon with energy \(E_{1}\) has momentum \(p_{1}\). If another photon has momentum \(p_{2}\) that is twice \(p_{1}\), how is the energy \(E_{2}\) of the second photon related to \(E_{1}\) ?
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