Max Frequency Homework: 9.0 ev Electron & 143 Neutrons, 92 Protons

  • Thread starter Thread starter aurao2003
  • Start date Start date
  • Tags Tags
    Frequency Maximum
Click For Summary
SUMMARY

The discussion focuses on solving two physics problems from the June 2009 Physics 1 paper, specifically regarding the maximum frequency of a photon emitted from a 9.0 eV electron and calculating the specific charge of an isotope with 143 neutrons and 92 protons. The maximum frequency is calculated using the equation f = E/h, yielding a frequency of approximately 2.17 x 1015 Hz. For the specific charge, the mass is determined by adding the number of protons and neutrons, resulting in a specific charge calculation of 92/(143+92) in electron charges per nucleon mass.

PREREQUISITES
  • Understanding of photon energy calculations using E=hf
  • Knowledge of specific charge formula: charge/mass
  • Familiarity with basic atomic structure, including protons and neutrons
  • Ability to perform unit conversions in physics calculations
NEXT STEPS
  • Research the concept of photon energy and frequency relationships in quantum mechanics
  • Study the calculation of specific charge for various isotopes
  • Explore the principles of atomic mass and its implications in nuclear physics
  • Learn about unit conversion techniques in scientific calculations
USEFUL FOR

Students studying physics, particularly those preparing for exams in quantum mechanics and nuclear physics, as well as educators looking for practical examples of energy and charge calculations.

aurao2003
Messages
122
Reaction score
0

Homework Statement



Hi
I am solving the june 2009 Physics 1 paper (Syllabus A). The mark scheme answer has got me slightly confused on 2 questions:

1. Calculate the maximum frequency of the photon emitted when the 9.0 ev
electron collides with an atom.

2.Calculate the specific charge of an isotope with 143 neutrons and 92 protons

Homework Equations


1) E= hf

2) Specific charge= charge/mass



The Attempt at a Solution



1)f= E/H
Therefore
f= 9 x 1.6x 10^-19/6.63 x 10^-34

But the schemes answer is very different to mine.

2)In the schemes answer, the mass was calculated as both 143 and 92. I presumed that the nucleon number constituted the mass, while the electrons (or number of protons) constituted the charge. (Multiplied by 92 in this case).

Anybody? Thanks.
 
Physics news on Phys.org
Your #1 calc looks good to me. Could it be some silly error in calculating? I got 2.17 x 10^15 Hz. What is the specified answer?

For #2, the mass is 143+92 (protons and neutrons have practically the same mass), so it is 92/(143+92) in electron charges per nucleon mass. Conversion of units may well be necessary.
 
Thanks. I am fine now.
 

Similar threads

Replies
9
Views
6K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 11 ·
Replies
11
Views
2K
Replies
27
Views
6K
Replies
4
Views
2K
Replies
1
Views
3K
Replies
22
Views
4K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 1 ·
Replies
1
Views
6K