AP EAMCET · Chemistry · Structure of Atom
The de Broglie wavelength of a particle of mass 1 mg moving with a velocity of \(10 \mathrm{~ms}^{-1}\) is (\(h=6.63 \times 10^{-34} \mathrm{~J} \mathrm{~s}\))
- A \(6.63 \times 10^{-29} \mathrm{~m}\)
- B \(6.63 \times 10^{-31} \mathrm{~m}\)
- C \(6.63 \times 10^{-34} \mathrm{~m}\)
- D \(6.63 \times 10^{-22} \mathrm{~m}\)
Answer & Solution
Correct Answer
(A) \(6.63 \times 10^{-29} \mathrm{~m}\)
Step-by-step Solution
Detailed explanation
According to de Broglie wavelength \((\lambda)=\frac{h}{p}\) or, \(\lambda=\frac{\mathrm{h}}{\mathrm{mv}}\) \(\left[\right.\) Where \(\mathrm{h}=\) Plank's constant \(=6.63 \times 10^{-34} \mathrm{~J}\) \(\mathrm{s}=\) momentum \(\mathrm{m}=\) mass of moving particle…
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