| ⇦ | 
| ⇨ |
The 6563 Å line emitted by hydrogen atom in a star is found to be red shifted by 5 Å. The speed with which the star is receding from the earth is
Options
(a) 17.29 x 10⁹ m/s
(b) 4.29 x 10⁷ m/s
(c) 3.39 x 10⁵ m/s
(d) 2.29 x 10⁵ m/s
Correct Answer:
2.29 x 10⁵ m/s
Explanation:
No explanation available. Be the first to write the explanation for this question by commenting below.
Related Questions: - In Young’s double silt experiment if the distance between the two slits
- The binding energy per nucleon in deuterium and helium nuclei are 1.1 MeV and 7.0 MeV
- The point of suspension λ of a simple pendulum with normal time period T₁ is moving
- Dimensional formula for electrical resistance R is given by
- A body is thrown upward from ground covers equal distances
Topics: Wave Optics
(101)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- In Young’s double silt experiment if the distance between the two slits
- The binding energy per nucleon in deuterium and helium nuclei are 1.1 MeV and 7.0 MeV
- The point of suspension λ of a simple pendulum with normal time period T₁ is moving
- Dimensional formula for electrical resistance R is given by
- A body is thrown upward from ground covers equal distances
Topics: Wave Optics (101)
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
question it self is wrong
THE CORRECT ONE WILL BE:
Q: The 6563A∘Hα6563A∘Hα line emitted by hydrogen in a star is found to be red shifted by 15∘A15∘A Estimate the speed with which the star is receding from the earth .
−6.86×105ms−16.86×105ms−1−6.86×10−5ms−16.86×10−5ms−1−6.86×105ms−16.86×105ms−1−6.86×10−5ms−16.86×10−5ms−1
Solution : given : λ=6563∘Aλ=6563∘A
Δλ=15∘AΔλ=15∘A
Δλ=−ΔλλΔλ=−Δλλcc
=−156563=−156563×3×108×3×108
=−6.86×105ms−1=−6.86×105ms−1
Answer : −6.86×105ms−1