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NEET PHYSICSMedium

The ionization energy of the electron in the hydrogen atom in its ground state is 13.6 eV. The atoms are excited to higher energy levels to emit radiations of 6 wavelengths. The maximum wavelength of emitted radiation corresponds to the transition between:

A

n=3 to n=2 states

B

n=3 to n=1 states

C

n=2 to n=1 states

D

n=4 to n=3 states

Step-by-Step Solution

This problem involves the Hydrogen Spectrum and Bohr's Model, covered in Unit 2: Atomic Structure (Chemistry) and Unit 18 (Physics). The sources confirm the energy levels and spectral series of hydrogen .

  1. Determine the excited state (nn): The number of spectral lines emitted when an electron drops from the nn-th state to the ground state is given by the formula n(n1)2\frac{n(n-1)}{2}. Given 6 wavelengths, n(n1)2=6n2n12=0(n4)(n+3)=0\frac{n(n-1)}{2} = 6 \Rightarrow n^2 - n - 12 = 0 \Rightarrow (n-4)(n+3) = 0. Thus, the electron is excited to the n=4n=4 state.
  2. Relate Wavelength to Energy: The wavelength (λ\lambda) of emitted radiation is inversely proportional to the energy difference of the transition (ΔE\Delta E), since ΔE=hν=hcλ\Delta E = h\nu = \frac{hc}{\lambda} . To find the maximum wavelength, we must identify the transition with the minimum energy difference.
  3. Identify the Transition: Energy levels in the hydrogen atom become closer together as nn increases (En1/n2E_n \propto -1/n^2) . Therefore, the smallest energy gap involving the excited state n=4n=4 is the transition to the immediate lower level, i.e., from n=4n=4 to n=3n=3. This corresponds to the first line of the Brackett series (infrared region) .
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