If the K.E of a free electron doubles, its de Broglie wavelength changes by the factor:
Correct answer: C. Option C
- A. Option A
- B. Option B
- C. Option C
- D. Option D
Explanation
The de Broglie wavelength of a particle is inversely related to its momentum, and momentum depends on the square root of kinetic energy. So, if the kinetic energy of a free electron doubles, its momentum increases by the square root of 2, and therefore its de Broglie wavelength decreases by a factor of 1 over √2.
Report an error
The more specific you are, the faster it gets fixed. A source beats an opinion.
Prefer email? [email protected]
About Atomic Spectra
Atoms emit or absorb light at specific wavelengths because electrons occupy quantized energy levels and change levels by absorbing or releasing photons. The topic covers emission and absorption spectra, spectral series, hydrogen’s line spectrum, energy-level transitions, and the relation between wavelength, frequency and photon energy.
Practise Atomic Spectra
356 free Atomic Spectra MCQs from Physics, each with the correct answer and an explanation. Unlimited attempts, no account needed.
Discussion
Stuck on an option, or know a faster way to get there? Ask or explain it here.
No comments yet. Be the first to explain this one.
Exams that ask Physics questions like this
Physics is on 16 papers prepared for on TestUstad, and all of them draw the same bank, so this question is worth knowing for every one of them.
More Atomic Spectra questions
For small uncertainty in the measurement of the position of a particle, the wavelength of the incident light should be:
If the series limit wavelength of the Lyman series 920 A0, then that for the Balmer series is:
The momentum of a photon is:
De-Broglie waves are associated with:
We can never accurately describe all aspects of subatomic particles simultaneously. It's correct according to:
If 13.6 eV energy is required to ionize the hydrogen atom, then the required energy to remove an electron from n = 2 is: