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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
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Subject: Chemistry
(2512)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- Co(NH₃)₅ONOCl₂ and Co(NH₃)₅NO₂Cl₂ show
- In the compounds KMnO₄ and K₂Cr₂O₇, the highest oxidation state is of the element
- What is formed when hydrogen atom of the -OH group in phenol is replaced by
- An alkane with a molecular formula C₆H₁₄ reacts with chlorine in the presence
- Which metal will displace copper from copper sulphate solution
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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