| ⇦ | 
| ⇨ |
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: - Calculate the amount of [Ca(OH)₂] required to remove the hardness of water
- Maximum bond angle at nitrogen is present in which of the following
- Roasting is done in
- The oxidation state of I in H₄IO₆⁻ is
- The solubility product of a sparingly soluble salt AB at room temperature
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
- Calculate the amount of [Ca(OH)₂] required to remove the hardness of water
- Maximum bond angle at nitrogen is present in which of the following
- Roasting is done in
- The oxidation state of I in H₄IO₆⁻ is
- The solubility product of a sparingly soluble salt AB at room temperature
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