Join the Most Relevant JEE Main 2025 Test Series & get 99+ percentile! Join Now
Search any question & find its solution
Question: Answered & Verified by Expert
$\mathrm{C}_{60}$ emerging from a source at a speed (v) has a de Broglie wavelength of $11.0 Å$. The value of $\mathrm{v}\left(\mathrm{inm} \mathrm{s}^{-1}\right.$ ) is closest to
[Planck's constant $\left.h=6.626 \times 10^{-34} \mathrm{~J} \mathrm{~s}\right]$
ChemistryStructure of AtomKVPYKVPY 2017 (19 Nov SA)
Options:
  • A $0.5$
  • B $2.5$
  • C $5.0$
  • D 30
Solution:
2010 Upvotes Verified Answer
The correct answer is: $0.5$
$\lambda=\frac{\mathrm{h}}{\mathrm{mv}} \Rightarrow \mathrm{v}=\frac{\mathrm{h}}{\mathrm{m} \cdot \lambda}=\frac{6.62 \times 10^{-34}}{720 \times 10^{-3} \times 11 \times 10^{-10}}$

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.