| ⇦ |
| ⇨ |
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 moment of inertia of a uniform circular disc of radius R and mass M about an axis
- The transition from the state 3 n = 3 to n = 1 in a hydrogen like atom results
- A juggler maintains four balls in motion,making each of them to rise
- The temperature at which the vapour pressure of a liquid becomes equal to the external
- In circular coil, when number of turns is doubled and resistance becomes 1/4th
Topics: Dual Nature of Matter and Radiation
(150)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- The moment of inertia of a uniform circular disc of radius R and mass M about an axis
- The transition from the state 3 n = 3 to n = 1 in a hydrogen like atom results
- A juggler maintains four balls in motion,making each of them to rise
- The temperature at which the vapour pressure of a liquid becomes equal to the external
- In circular coil, when number of turns is doubled and resistance becomes 1/4th
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.