Question 419: Atoms & Nuclei

Question

In hydrogen atom, transition from the state $\mathrm{n}=6$ to $n=1$ results in ultraviolet radiation. Infrared radiation will be obtained in the transition

Accepted Answer

Step 1: Recall hydrogen energy level formula
For hydrogen:
$ E_n = - \frac{13.6 \, 	ext{eV}}{n^2} $
The energy of the photon emitted in a transition $n_i 	o n_f$ is:
$ \Delta E = 13.6 \left(\frac{1}{n_f^2} - \frac{1}{n_i^2}ight) \, 	ext{eV} $
Step 2: Ultraviolet vs Visible vs Infrared
Large energy difference Ã¢ÂÂ higher frequency Ã¢ÂÂ shorter wavelength Ã¢ÂÂ ultraviolet.
Moderate energy difference Ã¢ÂÂ visible.
Small energy difference Ã¢ÂÂ longer wavelength Ã¢ÂÂ infrared.
Step 3: Analyze options
(A) $n=3 	o n=1$
:
$\Delta E = 13.6 \left(1 - \frac{1}{9}ight) = 13.6 	imes \frac{8}{9} \approx 12.1 \, 	ext{eV}$.
This corresponds to wavelength $\lambda \approx 1240/12.1 \approx 102 \,	ext{nm}$.
That is deep
UV
. Not IR.
(B) $n=4 	o n=2$
:
$\Delta E = 13.6 \left(\frac{1}{4} - \frac{1}{16}ight) = 13.6 	imes \frac{3}{16} = 2.55 \,	ext{eV}$.
$\lambda \approx 1240/2.55 \approx 486 \,	ext{nm}$ Ã¢ÂÂ
visible (blue-green)
. Not IR.
(C) $n=6 	o n=2$
:
$\Delta E = 13.6 \left(\frac{1}{4} - \frac{1}{36}ight) = 13.6 	imes \frac{8}{36} \approx 3.02 \,	ext{eV}$.
$\lambda \approx 1240/3.02 \approx 410 \,	ext{nm}$.
Visible (violet)
. Not IR.
(D) $n=5 	o n=3$:
$\Delta E = 13.6 \left(\frac{1}{9} - \frac{1}{25}ight) = 13.6 	imes \frac{16}{225} = 0.967 \,	ext{eV}$.
$\lambda \approx 1240/0.967 \approx 1280 \,	ext{nm}$. That's
Infrared
.
Ã¢ÂÂ
Correct Answer: Option D ($n=5 	o n=3$)

Answer

$\mathrm{n}=3$ to $\mathrm{n}=1$

Answer

$\mathrm{n}=4$ to $\mathrm{n}=2$

Answer

$n=6$ to $n=2$

Answer

$n=5$ to $n=3$

Atoms & Nuclei

In hydrogen atom, transition from the state \mathrm{n}=6 to n=1 results in ultraviolet radiation. Infrared radiation will be obtained in the transition

Choose the correct answer: