Question 416: Atoms & Nuclei

Question

The ratio of energies of photons produced due to transition of electron of hydrogen atom from its (i) third to $2^{	ext {nd }}$ energy level and (ii) highest energy level to $3^{	ext {rd }}$ level is

Accepted Answer

Case (i):
Electron transition from $n=3 	o n=2$.
Photon energy:
$ E = 13.6 \left(\frac{1}{n_2^2} - \frac{1}{n_1^2}ight)\ 	ext{eV}. $
So,
$ E_{(3 	o 2)} = 13.6 \left(\frac{1}{2^2} - \frac{1}{3^2}ight). $
$ = 13.6 \left(\frac{1}{4} - \frac{1}{9}ight). $
$ = 13.6 \left(\frac{9 - 4}{36}ight). $
$ = 13.6 \cdot \frac{5}{36}. $
$ = \frac{68}{36} = \frac{17}{9}\ 	ext{eV}. $
Case (ii):
Transition from highest energy level ($n = \infty$) to $n=3$.
Energy of photon:
$ E = 13.6 \left(\frac{1}{3^2} - \frac{1}{\infty^2}ight). $
$ = 13.6 \cdot \frac{1}{9}. $
$ = \frac{13.6}{9} = \frac{136}{90} = \frac{68}{45}\ 	ext{eV}. $
Now ratio:
$ \frac{E_{(3 	o 2)}}{E_{(\infty 	o 3)}} = \frac{	frac{17}{9}}{	frac{68}{45}}. $
Simplify:
$ = \frac{17}{9} \cdot \frac{45}{68}. $
$ = \frac{17 \cdot 45}{9 \cdot 68}. $
$ = \frac{17 \cdot 5}{68}. $
$ = \frac{85}{68}. $
$ = \frac{85}{68}. $
Divide numerator and denominator by 17:
$ = \frac{5}{4}. $
Final Answer:
$ \boxed{5:4} $
That matches
Option B
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Answer

$3: 2$

Answer

$5: 4$

Answer

$5: 3$

Answer

$8: 3$

Atoms & Nuclei

The ratio of energies of photons produced due to transition of electron of hydrogen atom from its (i) third to 2^{\text {nd }} energy level and (ii) highest energy level to 3^{\text {rd }} level is

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