Chapter 25: Problem 38
The work function of tungsten and sodium are \(4.5 \mathrm{eV}\) and \(2.3 \mathrm{eV}\) respectively. If the threshold wavelength, \(\lambda\) for sodium is \(5460 \AA\), the value of \(\lambda\) for tungsten is (a) \(2791 \dot{\mathrm{A}}\) (b) \(3260 \dot{A}\) (c) \(1925 \mathrm{~A}\) (d) \(1000 \mathrm{~A}\)
Short Answer
Step by step solution
Understanding Work Function and Threshold Wavelength
Relating Work Function and Threshold Wavelength for Sodium
Solving for Tungsten's Threshold Wavelength
Calculation and Comparison
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Work Function
- In our example, tungsten has a work function of 4.5 eV, while sodium has a work function of 2.3 eV.
- Therefore, electrons are more tightly bound to tungsten compared to sodium, requiring more energy to be emitted.
Threshold Wavelength
- Remember, as wavelength increases, the energy of the light decreases, because energy and wavelength are inversely related.
- Using the formula \[ \phi = \frac{hc}{\lambda} \]where \( \phi \) is the work function, we can calculate the threshold wavelength for different materials.
Planck's Constant
- Planck's constant \( h \) is used in the equation \[ E = h \times u \]which correlates energy \( E \) with frequency \( u \).
- It also appears in the formula \[ \phi = \frac{hc}{\lambda} \]where it helps to connect the work function \( \phi \) with the threshold wavelength \( \lambda \).