A hydrogen atom, initially in the ground stat | vedclass

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(C) The energy of the absorbed photon is given by $E = \frac{hc}{\lambda} = \frac{12500 \ 	ext{eV-\AA}}{980 \ 	ext{\AA}} \approx 12.755 \ 	ext{eV}$

Vedclass · Accepted Answer

$16$

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(C) The energy of the absorbed photon is given by $E = \frac{hc}{\lambda} = \frac{12500 \ 	ext{eV-\AA}}{980 \ 	ext{\AA}} \approx 12.755 \ 	ext{eV}$.The energy of the hydrogen atom in the ground state $(n=1)$ is $E_1 = -13.6 \ 	ext{eV}$.Let the atom be excited to an energy level $n$. The energy supplied is $E = E_n - E_1$.$12.755 = -\frac{13.6}{n^2} - (-13.6) = 13.6 \left(1 - \frac{1}{n^2}ight)$.$1 - \frac{1}{n^2} = \frac{12.755}{13.6} \approx 0.9378$.$\frac{1}{n^2} = 1 - 0.9378 = 0.0622$.$n^2 = \frac{1}{0.0622} \approx 16$.Thus, $n = 4$.The radius of the $n$-th orbit is given by $r_n = n^2 a_0$.For $n = 4$, $r_4 = 4^2 a_0 = 16 \ a_0$.

A hydrogen atom, initially in the ground state, is excited by absorbing a photon of wavelength $980 \ \text{\AA}$. The radius of the atom in the excited state, in terms of Bohr radius $a_0$, will be (given $hc = 12500 \ \text{eV-\AA}$). (in $a_0$)

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