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(C) The number of spectral lines emitted when an electron transitions from an excited state $n$ to the ground state is given by the formula $N = \frac

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$n=4 	o n=3$

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(C) The number of spectral lines emitted when an electron transitions from an excited state $n$ to the ground state is given by the formula $N = \frac{n(n-1)}{2}$.Given $N = 6$,we have $\frac{n(n-1)}{2} = 6$,which implies $n^2 - n - 12 = 0$. Solving this quadratic equation,we get $(n-4)(n+3) = 0$,so $n = 4$.The energy of the emitted photon is given by $\Delta E = E_i - E_f = \frac{hc}{\lambda}$.Since $\lambda = \frac{hc}{\Delta E}$,the wavelength $\lambda$ is maximum when the energy difference $\Delta E$ is minimum.The possible transitions from $n=4$ are: $(4 	o 3), (4 	o 2), (4 	o 1), (3 	o 2), (3 	o 1), (2 	o 1)$.Comparing the energy gaps,the transition $n=4 	o n=3$ has the smallest energy difference,and therefore corresponds to the maximum wavelength of the emitted radiation.

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

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