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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
- The oxidation states of Mn in MnSO₄,MnO₂,K₂MnO₄ and KMnO₄ are respectively
- Oxidation of acetaldehyde with selenium dioxide produces
- Which one of the following statements is not true
- Which of the following carbonyls will have the strongest C – O bond
- The reaction between an alkyl halide and an alkoxide ion to give an ether is
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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