/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 69 In his explanation of the thresh... [FREE SOLUTION] | 91Ó°ÊÓ

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In his explanation of the threshold frequency in the photoelectric effect, Einstein reasoned that the absorbed photon must have a minimum energy to dislodge an electron from the metal surface. This energy is called the work function (\phi) of the metal. What is the longest wavelength of radiation (in \(\mathrm{nm}\) ) that could cause the photoelectric effect in each of these metals: (a) calcium, \(\phi=4.60 \times 10^{-19}\) \(\mathrm{J} ;\) (b) titanium, \(\phi=6.94 \times 10^{-19} \mathrm{~J} ;\) (c) sodium, \(\phi=4.41 \times 10^{-19} \mathrm{~J} ?\)

Short Answer

Expert verified
Calcium: 432 nm, Titanium: 286 nm, Sodium: 451 nm.

Step by step solution

01

Understand the relationship between energy and wavelength

The energy of a photon is related to its wavelength by the equation: \[E = \frac{hc}{\lambda}\] where \(E\) is the energy of the photon, \(h\) is Planck's constant \(6.626 \times 10^{-34} \text{Js}\), \(c\) is the speed of light \(3.00 \times 10^8 \text{m/s}\), and \(\lambda\) is the wavelength of the photon.
02

Rearrange the formula to solve for wavelength

We need to find the wavelength, so rearrange the equation to solve for \(\lambda\): \[ \lambda = \frac{hc}{E} \]
03

Calculate the wavelength for calcium

For calcium, the work function (\( \phi \)) is given as \( 4.60 \times 10^{-19} \text{J} \). Use the formula to calculate \( \lambda \): \[ \lambda_{\text{Ca}} = \frac{6.626 \times 10^{-34} \text{Js} \times 3.00 \times 10^8 \text{m/s}}{4.60 \times 10^{-19} \text{J}} \approx 4.32 \times 10^{-7} \text{m} = 432 \text{nm} \]
04

Calculate the wavelength for titanium

For titanium, the work function (\( \phi \)) is given as \( 6.94 \times 10^{-19} \text{J} \). Use the formula to calculate \( \lambda \): \[ \lambda_{\text{Ti}} = \frac{6.626 \times 10^{-34} \text{Js} \times 3.00 \times 10^8 \text{m/s}}{6.94 \times 10^{-19} \text{J}} \approx 2.86 \times 10^{-7} \text{m} = 286 \text{nm} \]
05

Calculate the wavelength for sodium

For sodium, the work function (\( \phi \)) is given as \( 4.41 \times 10^{-19} \text{J} \). Use the formula to calculate \( \lambda \): \[ \lambda_{\text{Na}} = \frac{6.626 \times 10^{-34} \text{Js} \times 3.00 \times 10^8 \text{m/s}}{4.41 \times 10^{-19} \text{J}} \approx 4.51 \times 10^{-7} \text{m} = 451 \text{nm} \]

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Key Concepts

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

Threshold Frequency
In the photoelectric effect, the 'threshold frequency' is the minimum frequency of light required to eject electrons from the surface of a metal. When light hits the metal surface, it can cause electrons to be released, but only if the light carries enough energy, which is determined by its frequency. The energy of a photon is given by Planck's equation, which states that the energy is directly proportional to its frequency. This can be written as:

\[ E = h f \]

In this equation,
  • \(E\) is the energy of the photon,
  • \(h\) is Planck's constant
  • \(f\) is the frequency.
Light below this frequency doesn't have enough energy per photon to release electrons.

Understanding threshold frequency helps in knowing why certain wavelengths or frequencies of light can cause the photoelectric effect while others cannot.
Work Function
The 'work function' (\(\phi\)) is the minimum energy required to remove an electron from the surface of a metal. This is specific to each type of metal. The equation used here is:

\[ \phi = h f_{0} \]
Where
  • \(\phi\) is the work function
  • \(f_{0}\) is the threshold frequency
  • \(h\) is Planck's constant.
This concept is crucial because it tells us how much energy a photon must have to displace an electron. If the energy provided to the electron is less than the work function, the electron remains bonded to the metal.
Photon Energy
A photon is a particle of light that carries energy. The energy of a photon is calculated using the equation:

\[E = h f\]
where
  • \(E\) is the energy
  • \(h\) is Planck's constant
  • \(f\) is the frequency of the light.
This equation shows that the energy of a photon is directly proportional to its frequency. Higher frequency means higher energy and vice versa. Photon energy is a key player in understanding the photoelectric effect because only photons with enough energy can eject electrons from the metal surface.
Wavelength Calculation
The wavelength of light is inversely related to its frequency. One can calculate the wavelength if the energy or frequency of the photon is known using the equation:

\[\lambda = \frac{hc}{E} \]
where
  • \(\lambda\) is the wavelength
  • \(h\) is Planck's constant
  • \(c\) is the speed of light
  • \(E\) is the energy.
For example, given the work function (\(\phi\)) of calcium as \(4.60 \times 10^{-19} \text{J}\), the longest wavelength of light that can cause the photoelectric effect can be calculated. Rearranging the formula, we get the corresponding wavelength:

\[ \lambda = \frac{ (6.626 \times 10^{-34} \text{Js}) \times (3.00 \times 10^{8} \text{m/s}) }{ 4.60 \times 10^{-19} \text{J} } \approx 432 \text{nm} \]
Similar calculations can be done for other metals like titanium and sodium by using their respective work functions.
Planck's Constant
Planck's constant \((h)\) is a fundamental constant in physics. It has a value of \(6.626 \times 10^{-34}\) \text{Js}. Planck's constant plays a crucial role in quantum mechanics, particularly in the equations governing the photoelectric effect.

It relates the energy of a photon to its frequency:

\[ E = h f \]
Where
  • \(E\) is the energy
  • \(h\) is Planck's constant
  • \(f\) is the frequency.
In problems involving the photoelectric effect, Planck's constant helps us calculate the energy of photons and understand their interaction with electrons in metals. It's a key piece to linking macroscopic observations with microscopic behavior, showing how individual photons influence the emission of electrons from a metal surface.

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Most popular questions from this chapter

The quantum-mechanical treatment of the hydrogen atom gives this expression for the wave function, \(\psi,\) of the \(1 s\) orbital: $$ \Psi=\frac{1}{\sqrt{\pi}}\left(\frac{1}{a_{0}}\right)^{3 / 2} e^{-r l a_{0}} $$ where \(r\) is the distance from the nucleus and \(a_{0}\) is \(52.92 \mathrm{pm}\). The probability of finding the electron in a tiny volume at distance \(r\) from the nucleus is proportional to \(\psi^{2}\). The total probability of finding the electron at all points at distance \(r\) from the nucleus is proportional to \(4 \pi r^{2} \psi^{2}\). Calculate the values (to three significant figures) of \(\psi, \psi^{2},\) and \(4 \pi r^{2} \psi^{2}\) to fill in the following table and sketch a plot of each set of values versus \(r .\) Compare the latter two plots with those in Figure \(7.17 \mathrm{~A}\).

Horticulturists know that, for many plants, dark green leaves are associated with low light levels and pale green with high levels. (a) Use the photon theory to explain this behavior. (b) What change in leaf composition might account for this difference?

Use the Rydberg equation to find the wavelength (in \(\mathrm{nm}\) ) of the photon emitted when an electron in an \(\mathrm{H}\) atom undergoes a transition from \(n=5\) to \(n=2\)

Only certain transitions are allowed from one energy level to another. In one- electron species, the change in \(I\) for an allowed transition is \(\pm 1 .\) For example, a \(3 p\) electron can move to a \(2 s\) orbital but not to a \(2 p\). Thus, in the UV series, where \(n_{\text {final }}=1\), allowed transitions can start in a \(p\) orbital \((l=1)\) of \(n=2\) or higher, not in an \(s(l=0)\) or \(d(l=2)\) orbital of \(n=2\) or higher. From what orbital do each of the allowed transitions start for the first four emission lines in the visible series \(\left(n_{\text {final }}=2\right) ?\)

A sodium flame has a characteristic yellow color due to emission of light of wavelength \(589 \mathrm{nm}\). What is the mass equivalence of one photon with this wavelength \(\left(1 \mathrm{~J}=1 \mathrm{~kg} \cdot \mathrm{m}^{2} / \mathrm{s}^{2}\right) ?\)

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