A hydrogen atom is in the ground state. In or | vedclass

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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

Vedclass · Accepted Answer

$970$

Vedclass · Answer

(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,$(n-4)(n+3) = 0$,we get $n = 4$ (since $n$ must be positive).To excite the electron from the ground state $(n=1)$ to the $n=4$ state,the energy of the incident photon must be $\Delta E = E_4 - E_1 = 13.6 \left(1 - \frac{1}{4^2}ight) 	ext{ eV}$.$\Delta E = 13.6 	imes \left(1 - \frac{1}{16}ight) = 13.6 	imes \frac{15}{16} = 12.75 	ext{ eV}$.The wavelength $\lambda$ is given by $\lambda = \frac{12400 	ext{ eV } \mathring A}{12.75 	ext{ eV}} \approx 972.5 \mathring A$.Rounding to the nearest provided option,the correct answer is $970 \mathring A$.

$A$ hydrogen atom is in the ground state. In order to get six lines in its emission spectrum,the wavelength of the incident radiation should be .......... $\mathring A$.

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