The momentum of a particle having a de Broglie wavelength of 10⁻¹⁷metres is

⇦

⇨

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:

1. Which of the following substances will give amide, when reacted with NH₃
2. When conc. H₂SO₄ is added to dry KNO₃,brown fumes are evolved.These fumes are
3. Magnesium carbonate of x g decomposes. What is the percentage of purity
4. Formula of ferric sulphate is
5. Two different gases enclosed in different flasks A and B at same temperature and pressure

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