Question 381: Atoms & Nuclei

Question

Out of the following transitions in hydrogen atom, identify the transition which emits photons of highest frequency.

Accepted Answer

Step 1: Recall the formula
The energy difference between two levels in the hydrogen atom is:
$ \Delta E = 13.6 \, 	ext{eV}\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}ight) $
(for $ n_2 > n_1 $).
The emitted photon frequency is directly proportional to this energy difference:
$ 
u = \frac{\Delta E}{h} $
So the larger the energy difference, the higher the frequency.
Step 2: Analyze each option
Option A: $ n=1 	o n=2 $
This describes excitation (absorption), not emission. Even if it did produce a photon, this is absorption, not emission.
Option B: $ n=2 	o n=1 $
This is emission. Energy difference:
$ \Delta E = 13.6\left(1 - \frac{1}{4}ight) = 13.6\left(\frac{3}{4}ight) = 10.2 \, 	ext{eV} $
Option C: $ n=2 	o n=6 $
This would be excitation (absorption), not emission.
Option D: $ n=6 	o n=2 $
This is emission. Energy difference:
$ \Delta E = 13.6\left(\frac{1}{4} - \frac{1}{36}ight)  = 13.6\left(\frac{9 - 1}{36}ight)  = 13.6 	imes \frac{8}{36}  = 13.6 	imes \frac{2}{9}  \approx 3.02 \, 	ext{eV} $
Step 3: Compare
$ n=2 	o n=1 $: $ \Delta E = 10.2 \, 	ext{eV} $
$ n=6 	o n=2 $: $ \Delta E = 3.02 \, 	ext{eV} $
So the photon from $ n=2 	o n=1 $ has higher frequency (larger energy).
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Final Answer:
$ 	extbf{Option B: } n=2 	o n=1 $

Answer

$\mathrm{n}=1$ to $\mathrm{n}=2$

Answer

$\mathrm{n}=2$ to $\mathrm{n}=1$

Answer

$n=2$ to $n=6$

Answer

$n=6$ to $n=2$

Atoms & Nuclei

Out of the following transitions in hydrogen atom, identify the transition which emits photons of highest frequency.

Choose the correct answer: