Modern Physics Ques 19
- 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) $25 a _0$
Show Answer
Solution:
Formula:
- 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, $r _n=n^{2} a _0 \Rightarrow r _n=16 a _0$
BETA
