| ⇦ |
| ⇨ |
If h is Planck’s constant, the momentum of a photon of wavelength 0.01 Å is
Options
(a) 10⁻² h
(b) h
(c) 10² h
(d) 10¹² h
Correct Answer:
10¹² h
Explanation:
No explanation available. Be the first to write the explanation for this question by commenting below.
Related Questions: - The energy released in the fission of 1 kg of ₉₂U²³⁵ is (energy per fission = 200 MeV)
- A body of weight W newton is at the surface of the earth.
- A ball is dropped into a well in which the water level is at a depth h below the top
- The current in the coil of inductance 5H decreases at the rate of 2 A/s
- When ₃Li⁷ nuclei are bombarded by protons, and the resultant nuclei are ₄Be⁸,
Topics: Dual Nature of Matter and Radiation
(150)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- The energy released in the fission of 1 kg of ₉₂U²³⁵ is (energy per fission = 200 MeV)
- A body of weight W newton is at the surface of the earth.
- A ball is dropped into a well in which the water level is at a depth h below the top
- The current in the coil of inductance 5H decreases at the rate of 2 A/s
- When ₃Li⁷ nuclei are bombarded by protons, and the resultant nuclei are ₄Be⁸,
Topics: Dual Nature of Matter and Radiation (150)
Subject: Physics (2479)
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

momentum is given by p=h/lambda
As h is constant so p is inversely proportional to lambda. The value of lambda is 0.01 angstrom I.e, 1×102 Angstrom. and angstrom is equals to 10 raised to the power 10. so the equation becomes p=1/1×102×1010. p=1/1×1012
p=10 raised to the power -12. So the answer is ×10-12.