Question

A hydrogen atom, initially in the ground state is excited by absorbing a photon of wavelength $980\stackrel{0}{A}$. The radius of the atom in the excited state, in terms of Bohr radius $a_0$, will be: $\left(h_c=12500eV-\stackrel{0}{A}\right)$

• A. $9a_0$
• B. $25a_0$
• C. $4a_0$
• D. $16a_0$

Answer 1

The correct option is D

$16a_0$

Step 1: Given data.

Wavelength of the photon $\lambda =980\stackrel{0}{A}$

$h_c=12500eV-\stackrel{0}{A}$

Step 2: Finding net charge in the energy of photon.

We know that, energy of photon in a transition with photon wavelength $\lambda$.

$∆E=\frac{h_c}{\lambda }$

$\Rightarrow$$∆E=\frac{12500}{980}$ ……$\left(i\right)$

Now,

$∆E=E_n-E_1$ ……..$\left(ii\right)$

Where, $E_n$​ is the energy of an electron in $n^{th}$ Bohr's orbit.

Therefore,

$E_n=-\frac{13.6}{n^2}eV$ …..$\left(iii\right)$

Thus for $E_1$, $n=1$

$E_1=-13.6eV$ …..$\left(iv\right)$

From equation $\left(i\right)$ and $\left(ii\right)$ we get.

$E_n-E_1=\frac{12500}{980}$

$\Rightarrow$$-\frac{13.6}{n^2}-\left(-13.6\right)=\frac{12500}{980}$ $\left[Fromequation\left(ii\right)and\left(iv\right)\right]$

$\Rightarrow$ $-\frac{13.6}{n^2}=12.76-13.6$

$\Rightarrow$ $n^2=\frac{13.6}{0.84}$

$\Rightarrow$ $n=4\left(approx\right)$ …..$\left(v\right)$

Step3: Finding radius of the ${\mathit{n}}^{\mathbf{t}\mathbf{h}}$orbit becomes.

We know that

$r_n=n^2{a}_0$

$\Rightarrow$$r_4=4^2{a}_0$ $\left[Fromequation\left(v\right)\right]$

$\Rightarrow$$r_4=16a_0$

Hence, option D is correct.