| ⇦ | 
| ⇨ |
An electron in the hydrogen atom jumps from excited state n to the ground state. The wavelength so emitted illuminates a photosensitive material having work function 2.75 eV. If the stopping potential of the photoelectron is 10 V, the value of n is
Options
(a) 3
(b) 4
(c) 5
(d) 2
Correct Answer:
4
Explanation:
KEₘₐₓ = 10 eV ? = 2.75 eV
Total incident energy
E = ? + KEₘₐₓ = 12.75 eV
Energy is released when electron jumps from the excited state n to the ground state.
E₄ – E₁ = {-0.85 – (-13.6) eV}
= 12.75 eV value of n = 4.
Related Questions: - In Young’s double slit experiment with sodium vapour lamp of wavelength
- A linear aperture whose width is 0.02 cm is placed immediately in front of a lens
- Consider 3rd orbit of He⁺(helium), using non-relativistic approach, the speed
- Changing magnetic fields can set up current loops in nearby metal bodies
- The ionisation potential of hydrogen-atom is -13.6 eV. An electron in the groundstate
Topics: Atoms and Nuclei
(136)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- In Young’s double slit experiment with sodium vapour lamp of wavelength
- A linear aperture whose width is 0.02 cm is placed immediately in front of a lens
- Consider 3rd orbit of He⁺(helium), using non-relativistic approach, the speed
- Changing magnetic fields can set up current loops in nearby metal bodies
- The ionisation potential of hydrogen-atom is -13.6 eV. An electron in the groundstate
Topics: Atoms and Nuclei (136)
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
Leave a Reply