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Question

(B) The number of spectral lines emitted when an electron transitions from an excited state $n_2$ to the ground state $n_1=1$ is given by the formula

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(B) The number of spectral lines emitted when an electron transitions from an excited state $n_2$ to the ground state $n_1=1$ is given by the formula $N = \frac{n_2(n_2-1)}{2}$.Given $N = 6$,we have $\frac{n_2(n_2-1)}{2} = 6$,which implies $n_2^2 - n_2 - 12 = 0$.Solving this quadratic equation,we get $(n_2-4)(n_2+3) = 0$. Since $n_2 > 0$,we find $n_2 = 4$.The energy of the incident photon must be equal to the energy difference between the ground state $(n=1)$ and the excited state $(n=4)$:$h
u = E_4 - E_1 = -0.85 \ eV - (-13.6 \ eV) = 12.75 \ eV$.Given $h = 4.25 	imes 10^{-15} \ eVs$,the frequency $
u$ is:$
u = \frac{12.75 \ eV}{4.25 	imes 10^{-15} \ eVs} = 3 	imes 10^{15} \ Hz$.Comparing this with $x 	imes 10^{15} \ Hz$,we get $x = 3$.

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