Modern Physics 1 Question 12

9. A hydrogen atom, initially in the ground state is excited by absorbing a photon of wavelength $980 \AA$. The radius of the atom in the excited state in terms of Bohr radius $a _0$ will be (Take $h c=12500 eV-\AA)$

(Main 2019, 11 Jan I)

(a) $4 a _0$

(b) $9 a _0$

(c) $16 a _0$

(d) $25 a _0$

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Solution:

  1. We know that net change in energy of a photon in a transition with wavelength $\lambda$ is $\Delta E=h c / \lambda$.

Here, $h c=12500 eV \AA$ and $\lambda=980 \AA$

$$ \begin{aligned} \Delta E & =12500 / 980=12.76 eV \\ \Rightarrow \quad E _n-E _1 & =12.76 eV \end{aligned} $$

Since, the energy associated with an electron in $n^{\text {th }}$ Bohr’s orbit is given as,

$$ \begin{aligned} E _n & =\frac{-13.6}{n^{2}} eV \\ E _n & =E _1+12.76 eV \\ & =\frac{-13.6}{(1)^{2}}+12.76=-0.84 \end{aligned} $$

Putting this value in Eq. (i)

$$ \Rightarrow \quad n^{2}=\frac{-13.6}{-0.84}=16 \Rightarrow n=4 $$

and radius of $n^{\text {th }}$ orbit, $r _n=n^{2} a _0 \Rightarrow r _n=16 a _0$



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