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In hydrogen spectrum, the third line fro...

In hydrogen spectrum, the third line from the red end corresponds to which one of the following inter-orbit jumps of the electron for Bohr orbits in an atom of hydrogen ?

A

`5 rarr 2`

B

`4 rarr 1`

C

`2 rarr 5`

D

`3 rarr 2`

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The correct Answer is:
To determine the inter-orbit jump of the electron in the hydrogen atom corresponding to the third line from the red end in the hydrogen spectrum, we will follow these steps: ### Step 1: Understand the Series The hydrogen spectrum consists of several series, and the visible lines correspond to the Balmer series. The Balmer series involves transitions where the electron falls to the n=2 energy level from higher energy levels (n=3, 4, 5, etc.). **Hint:** Remember that the Balmer series is the only series that produces visible light. ### Step 2: Identify the Ground State In the Balmer series, the ground state (final state) is fixed at n=2. The initial state (n2) can be any value greater than 2 (n=3, 4, 5, ...). **Hint:** The ground state for the Balmer series is always n=2. ### Step 3: Determine the Line Position The lines in the Balmer series correspond to different transitions: - The first line (H-alpha) corresponds to the transition from n=3 to n=2. - The second line (H-beta) corresponds to the transition from n=4 to n=2. - The third line (H-gamma) corresponds to the transition from n=5 to n=2. **Hint:** Count the lines starting from the red end (longest wavelength) to identify the transitions. ### Step 4: Identify the Third Line Since we are looking for the third line from the red end, we need to find the transition that corresponds to this line. As established: - 1st line: n=3 to n=2 (H-alpha) - 2nd line: n=4 to n=2 (H-beta) - 3rd line: n=5 to n=2 (H-gamma) Thus, the third line corresponds to the transition from n=5 to n=2. **Hint:** The third line corresponds to the highest energy level that transitions to n=2. ### Final Answer The third line from the red end in the hydrogen spectrum corresponds to the electron transition from n=5 to n=2. **Correct Option:** 5 to 2 (Option 1).

To determine the inter-orbit jump of the electron in the hydrogen atom corresponding to the third line from the red end in the hydrogen spectrum, we will follow these steps: ### Step 1: Understand the Series The hydrogen spectrum consists of several series, and the visible lines correspond to the Balmer series. The Balmer series involves transitions where the electron falls to the n=2 energy level from higher energy levels (n=3, 4, 5, etc.). **Hint:** Remember that the Balmer series is the only series that produces visible light. ### Step 2: Identify the Ground State ...
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PRADEEP-STRUCTURE OF ATOM-Competition Focus (JEE (Main and Advanced)/Medical Entrance (I. Multiple Choice Question) With one correct Answer
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  2. The angular momentum of electron in 'd' orbital is equal to:

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  3. In hydrogen spectrum, the third line from the red end corresponds to w...

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  4. The frequency of radiation emiited when the electron falls n =...

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  5. Energy of an electron is given by E = - 2.178 xx 10^-18 J ((Z^2)/(n^2)...

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  6. the radius of which of the following orbit is same as that of the firs...

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  7. One electron species having ionization enegry of 54.4 eV is

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  8. The ratio of area covered by second orbital to the first orbital is.

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  9. The kinetic energy of an electron in the second Bohr orbit of a hydrog...

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  10. The most probable radius (in pm) for finding the electron in He^(+) is

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  11. If Delta E is the energy emitted in electron volts when an electronic ...

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  12. If R(H) represents Rydberg constant, then the energy of the electron i...

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  13. Ionisation energy of He^+ is 19.6 xx 10^-18 J "atom"^(-1). The energy ...

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  14. If an electron travels with a velocity of 1/100th speed of light in th...

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  15. In the hydrogen atom, the electrons are excited to the 5th energy leve...

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  16. Which of the following is the energy of a possible excited state of h...

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  17. Calculate the energy in joule corresponding to light of wavelength 45 ...

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  18. The radius of the second Bohr orbit for hydrogen atom is : (Planck's...

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  19. Schrodinger wave equation for a particle in a one-dimension box is

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  20. Psi^(2) = 0 represents

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