| ⇦ | 
| ⇨ |
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.
Related Questions: - Which of the following is isoelectronic?
- The Formation of the oxide ion, O²⁻ from oxygen atom requires first an exothermic
- Hydrolysis of an ester in an alkaline medium is
- How much quick lime can be obtained from 25g of CaCO₃?
- 6.4 g SO₂ at 0⁰C and 0.99 atm pressure occupies a volume of 2.24 L
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
- Which of the following is isoelectronic?
- The Formation of the oxide ion, O²⁻ from oxygen atom requires first an exothermic
- Hydrolysis of an ester in an alkaline medium is
- How much quick lime can be obtained from 25g of CaCO₃?
- 6.4 g SO₂ at 0⁰C and 0.99 atm pressure occupies a volume of 2.24 L
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
18000+ students are using NEETLab to improve their score. What about you?
Solve Previous Year MCQs, Mock Tests, Topicwise Practice Tests, Identify Weak Topics, Formula Flash cards and much more is available in NEETLab Android App to improve your NEET score.
Share this page with your friends
Leave a Reply