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The wave number of the first Balmer line of Li2+ ion is 1, 36, 800 cm-1. The wave number of the first line of Balmer series of hydrogen atom is (in cm-1)
  • a)
    68,400
  • b)
    15,200
  • c)
    76,000
  • d)
    30,800
Correct answer is option 'B'. Can you explain this answer?
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The wave number of the first Balmer line of Li2+ ion is 1, 36, 800 cm-...
Given Data:
Wave number of the first Balmer line of Li2+ ion = 1,36,800 cm^-1

Formula:
The formula for calculating the wave number of the first line of the Balmer series for any atom is given by:
\[ \tilde{\nu} = R_H \left( \dfrac{1}{n_f^2} - \dfrac{1}{n_i^2} \right) \]
where,
- \( \tilde{\nu} \) = Wave number
- \( R_H \) = Rydberg constant for hydrogen atom
- \( n_f \) = final energy level
- \( n_i \) = initial energy level

Solution:
We are given that the wave number of the first Balmer line of Li2+ ion is 1,36,800 cm^-1. This corresponds to the transition from the second energy level to the first energy level.
Now, we need to find the wave number of the first line of the Balmer series for hydrogen atom.
Substitute the values in the formula:
For Li2+ ion: \( n_f = 1 \) and \( n_i = 2 \)
\[ 1,36,800 = R_H \left( \dfrac{1}{1^2} - \dfrac{1}{2^2} \right) \]
\[ 1,36,800 = R_H \left( 1 - \dfrac{1}{4} \right) \]
\[ 1,36,800 = R_H \times \dfrac{3}{4} \]
Now, to find the wave number for hydrogen atom:
For hydrogen atom: \( n_f = 2 \) and \( n_i = 3 \)
\[ \tilde{\nu} = R_H \left( \dfrac{1}{2^2} - \dfrac{1}{3^2} \right) \]
\[ \tilde{\nu} = R_H \left( \dfrac{1}{4} - \dfrac{1}{9} \right) \]
\[ \tilde{\nu} = R_H \times \left( \dfrac{5}{36} \right) \]
Comparing the two expressions:
\[ 1,36,800 = R_H \times \dfrac{3}{4} \]
\[ \tilde{\nu} = R_H \times \dfrac{5}{36} \]
Therefore, by solving for \( R_H \), we can calculate the wave number of the first line of the Balmer series for hydrogen atom.
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The only electron in the hydrogen atom resides under ordinary conditions on the first orbit. When energy is supplied, the electron moves to higher energy orbit depending on the amount of energy absorbed. When this electron returns to any of the lower orbits, it emits energy. Lyman series is formed when the electron returns to the lowest orbit while Balmer series is formed when the electron returns to second orbit. Similarly, Paschen, Brackett and Pfund series are formed when electron returns to the third, fourth and fifth orbits from higher energy orbits respectively. Maximum number of lines produced when an electron jumps from nth level to ground level is equal to . For examle, in the case of n = 4, number of lines produced is 6. (4 3, 4 2, 4 1, 3 2, 3 1, 2 1). When an electron returns from n2 to n1 state, the number of lines in the spectrum will be equal toIf the electron comes back from energy level having energy E2 to energy level having energy E1, then the difference may be expressed in terms of energy of photon as :Since h and c are constants, E corresponds to definite energy; thus each transition from one energy level to another will produce a light of definite wavelength. This is actually observed as a line in the spectrum of hydrogen atom.Wave number of line is given by the formulawhere R is a Rydbergs constant(R = 1.1 107 m-1)Q.The difference in the wavelength of the 1st line of Lyman series and 2nd line of Balmer series in a hydrogen atom is

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The wave number of the first Balmer line of Li2+ ion is 1, 36, 800 cm-1. The wave number of the first line of Balmer series of hydrogen atom is (in cm-1)a)68,400b)15,200c)76,000d)30,800Correct answer is option 'B'. Can you explain this answer?
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