Search any question & find its solution
Question:
Answered & Verified by Expert
An electron is moving with an initial velocity \( \vec{V}=V_{0} \hat{i} \) and is in a uniform magnetic field
\( \vec{B}=B_{0} \hat{j} \). Then its de Broglie wavelength
Options:
\( \vec{B}=B_{0} \hat{j} \). Then its de Broglie wavelength
Solution:
1925 Upvotes
Verified Answer
The correct answer is:
remains constant
(B)
de-Broglie wavelength \( \lambda=\frac{h}{m v} \)
Electron is entering perpendicular to magnetic field. It moves in a circular path. But its speed remains same. Therefore
de Broglie wavelength also remains constant.
de-Broglie wavelength \( \lambda=\frac{h}{m v} \)
Electron is entering perpendicular to magnetic field. It moves in a circular path. But its speed remains same. Therefore
de Broglie wavelength also remains constant.
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.