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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
- Write the IUPAC name of CH₃CH₂COOH
- At 0K, (i) ¹²C and (ii) a mixture of ¹²C and ¹⁴C will
- The geometry of Ni(CO)₄ and Ni(PPh₃)₂Cl₂ are
- On reaction with Mg, very dilute nitric acid produces
- Electrophile in the case of chlorination of benzene in the presence of FeCl₃ 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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