Home
Class 11
CHEMISTRY
Assertion (A) : Limiting line is the bal...

Assertion (A) : Limiting line is the balmer series has a wavelength of `364.4 nm`
Reason (R ) : Limiting line is obtained for a jump electron from `n = infty`

A

Statement-I is true, Statement-II is true , Statement-II is correct explanation for Statement-I.

B

Statement-I is true, Statement-II is true , Statement-II is NOT a correct explanation for statement-I

C

Statement-I is true, Statement-II is false

D

Statement-I is false, Statement-II is true

Text Solution

AI Generated Solution

The correct Answer is:
To solve the question, we need to analyze both the assertion and the reason provided. ### Step 1: Understand the Assertion The assertion states that the limiting line of the Balmer series has a wavelength of 364.4 nm. The Balmer series corresponds to electronic transitions in a hydrogen atom where the electron falls to the n=2 energy level from higher energy levels (n > 2). ### Step 2: Identify the Limiting Line The limiting line in the Balmer series occurs when the electron transitions from an infinite energy level (n = ∞) to the n = 2 energy level. This transition corresponds to the maximum energy difference, resulting in the shortest wavelength in the series. ### Step 3: Use the Rydberg Formula To find the wavelength of the limiting line, we can use the Rydberg formula for hydrogen: \[ \frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \] Where: - \( R \) is the Rydberg constant (approximately \( 1.097 \times 10^7 \, \text{m}^{-1} \)) - \( n_1 = 2 \) (the final energy level for the Balmer series) - \( n_2 = \infty \) (the initial energy level for the limiting line) ### Step 4: Substitute Values into the Formula Substituting the values into the Rydberg formula: \[ \frac{1}{\lambda} = R \left( \frac{1}{2^2} - \frac{1}{\infty^2} \right) \] Since \( \frac{1}{\infty^2} = 0 \): \[ \frac{1}{\lambda} = R \left( \frac{1}{4} \right) \] ### Step 5: Solve for Wavelength Now, we can express the wavelength \( \lambda \): \[ \lambda = \frac{4}{R} \] Substituting the value of \( R \): \[ \lambda = \frac{4}{1.097 \times 10^7} \approx 364.4 \, \text{nm} \] ### Step 6: Conclusion Thus, the assertion is correct as the limiting line of the Balmer series indeed has a wavelength of 364.4 nm. The reason is also correct because the limiting line is obtained when an electron jumps from \( n = \infty \) to \( n = 2 \). ### Final Answer Both the assertion and the reason are correct, and the reason correctly explains the assertion. ---
Promotional Banner

Topper's Solved these Questions

  • ATOMIC STRUCTURE

    ALLEN|Exercise Exercise - 04[A]|49 Videos
  • ATOMIC STRUCTURE

    ALLEN|Exercise Exercise - 04[B]|10 Videos
  • ATOMIC STRUCTURE

    ALLEN|Exercise Exercise - 02|59 Videos
  • IUPAC NOMENCLATURE

    ALLEN|Exercise Exercise - 05(B)|7 Videos

Similar Questions

Explore conceptually related problems

Assertion (A) : Limiting line is the balmer series ghas a wavelength of 364.4 nm Reason (R ) : Limiting line is obtained for a jump electyron from n = prop

The limiting line Balmer series will have a frequency of

The limiting line Balmer series will have a frequency of

The wavelength of the first line in the balmer series is 656 nm .Calculate the wavelength of the second line and the limiting line in the balmer series

A line in Lyman series of hydrogen atom has a wavelength of 1.03 xx 10^(-7)m . What is the initial energy level of the electron ?

Wavelength of the first line of balmer seris is 600 nm. The wavelength of second line of the balmer series will be

Find the ratio of series limit wavelength of Balmer series to wavelength of first time line of paschen series.

Assertion (A) : Each electron in an atom has two spin quantum number Reason (R ) : Spin quantum numbers are obtained by solving schrodinger wave equation

Calcualte the limiting wavelengths of Balmer Series. If wave length of first line of Balmer series is 656 nm.

In hydrogen spectrum L_(alpha) line arises due to transition of electron from the orbit n=3 to the orbit n=2. In the spectrum of singly ionized helium there is a line having the same wavelength as that of the L_(alpha) line. This is due to the transition of electron from the orbit:

ALLEN-ATOMIC STRUCTURE-Exercise - 03
  1. {:(,"Column-I",,"Column-II",),((A),"Cathode rays",(p),"Helium nuclei",...

    Text Solution

    |

  2. Frequancy =f(1), Time period = T, Energy of n^(th) orbit = E(n), radiu...

    Text Solution

    |

  3. Statement-I : Nodal plane of p(x) atmoic orbital is yz plane. Becau...

    Text Solution

    |

  4. Statement-I : No two electrons in an atom can have the same values of ...

    Text Solution

    |

  5. Assertion (A) : p orbital is dumb- bell shaped Reason (R ) :Electron...

    Text Solution

    |

  6. Statement-I : The ground state configuration of Cr is 3d^(5) 4s^(1). ...

    Text Solution

    |

  7. Statement-I : Mass numbers of most of the elements are fractional. ...

    Text Solution

    |

  8. Assertion (A) : Limiting line is the balmer series has a wavelength of...

    Text Solution

    |

  9. Assertion (A) : The electronic configuration of nitrogen atom is repre...

    Text Solution

    |

  10. Assertion:The configuration of boron atom cannot be 1s^2 2s^3 Reason...

    Text Solution

    |

  11. Statement-I : 2p orbitals do not have spherical nodes. Because S...

    Text Solution

    |

  12. Statement-I : In Rutherford's gold foil experiment, very few alpha-par...

    Text Solution

    |

  13. Statement-I : Each electron in an atom has two spin quantum numbers. ...

    Text Solution

    |

  14. Statement-I : There are two spherical nodes in 3s-orbital. Because ...

    Text Solution

    |

  15. Comprehension # 1 Read the following rules and answer the questions ...

    Text Solution

    |

  16. Comprehension # 1 Read the following rules and answer the questions ...

    Text Solution

    |

  17. Comprehension # 1 Read the following rules and answer the questions ...

    Text Solution

    |

  18. Comprehension # 1 Read the following rules and answer the questions ...

    Text Solution

    |

  19. Comprehension # 1 Read the following rules and answer the questions ...

    Text Solution

    |

  20. Comprehension # 1 Read the following rules and answer the questions ...

    Text Solution

    |