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Question

(C) The energy of a photon emitted during a transition is given by $h
u = E_n - E_m$.For the series limit,the electron transitions from $n = \infty$

Vedclass · Accepted Answer

$\frac{v_L}{25}$

Vedclass · Answer

(C) The energy of a photon emitted during a transition is given by $h
u = E_n - E_m$.For the series limit,the electron transitions from $n = \infty$ to the ground state of the series.For the Lyman series,the ground state is $n_1 = 1$. Thus,$h
u_L = E_{\infty} - E_1 = 0 - E_1 = -E_1$.For the Pfund series,the ground state is $n_5 = 5$. Thus,$h
u_f = E_{\infty} - E_5 = 0 - E_5 = -E_5$.Since $E_n = \frac{E_1}{n^2}$,we have $E_5 = \frac{E_1}{5^2} = \frac{E_1}{25}$.Substituting this into the expression for $
u_f$:$h
u_f = -\left(\frac{E_1}{25}ight) = \frac{-E_1}{25}$.Since $h
u_L = -E_1$,we get $h
u_f = \frac{h
u_L}{25}$.Therefore,$
u_f = \frac{
u_L}{25}$.

If the series limit frequency of the Lyman series is $v_L$,then the series limit frequency of the Pfund series is

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