Question

The energy of first (lowest) Balmer line of H atom is \(x\) J. The energy (in J) of second Balmer line of H atom is:

• A. \(\dfrac{x}{1.35}\)
• B. \(x^2\)
• C. \(1.35x\)
• D. \(2x\)

Hint:

Higher Balmer lines have greater transition energy because the initial energy level is higher.

Answer 1

The Correct Option is C

The problem involves finding the energy of the second Balmer line of the hydrogen atom, given the energy of the first Balmer line.

The Balmer series corresponds to electronic transitions in which an electron falls to the second energy level (\( n_2 = 2 \)) from a higher level (\( n_1 > 2 \)).

Energy of a Balmer Transition

The energy of the photon emitted during a transition is given by:

\[ E = R_H \left( \frac{1}{n_2^2} - \frac{1}{n_1^2} \right) \]

where:

• \(R_H = 2.18 \times 10^{-18}\,\text{J}\) (Rydberg constant)
• \(n_2 = 2\)
• \(n_1\) is the initial energy level 

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First Balmer Line

For the first Balmer line, \( n_1 = 3 \):

\[ E_1 = R_H \left( \frac{1}{2^2} - \frac{1}{3^2} \right) \]

Second Balmer Line

For the second Balmer line, \( n_1 = 4 \):

\[ E_2 = R_H \left( \frac{1}{2^2} - \frac{1}{4^2} \right) \]

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Ratio of Energies

The ratio of the energies is:

\[ \frac{E_2}{E_1} = \frac{\left( \frac{1}{4} - \frac{1}{16} \right)} {\left( \frac{1}{4} - \frac{1}{9} \right)} \]

\[ = \frac{\frac{3}{16}}{\frac{5}{36}} = \frac{108}{80} = 1.35 \]

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Final Answer

Thus, the energy of the second Balmer line is 1.35 times the energy of the first Balmer line.

\[ \boxed{1.35x} \]