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The momentum of a particle having a de Broglie wavelength of 10⁻¹⁷metres is (Given h= 6.625 × 10⁻³⁴Js)
Options
(a) 3.3125 x 10⁻⁷kg ms⁻¹
(b) 26.5 x 10⁻⁷kg ms⁻¹
(c) 6.625 x 10⁻¹⁷kg ms⁻¹
(d) 13.25 x 10⁻¹⁷kg ms⁻¹
Correct Answer:
6.625 x 10⁻¹⁷kg ms⁻¹
Explanation:
According to de broglie λ=h/mv ⇒ mv= h/λ = 6.626 x 10⁻³⁴ /10⁻¹⁷ = 6.626 x 10⁻¹⁷ kg m/s.
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Question Type: Memory
(964)
Difficulty Level: Easy
(1008)
Topics: Structure of Atom
(90)
Subject: Chemistry
(2512)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- Platinum, palladium and iridium are called noble metals because
- Considering the state of hybridization of carbon atoms, find out the molecule
- If active mass of a 6% solution of a compound is 2, its molecular weight will be
- Which of the following is the lightest metal
- In N₂ + 3H₂ → 2NH₃ reversible reaction ,increases in pressure will favour
Question Type: Memory (964)
Difficulty Level: Easy (1008)
Topics: Structure of Atom (90)
Subject: Chemistry (2512)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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