JEE Main · 2022 · 25 Jul 2022 · Shift 2 · Modern Physics

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\frac{x}{x + 4}

is the ratio of energies of photons produced due to transition of an electron of hydrogen atom from its
(i) third permitted energy level to the second level and
(ii) the highest permitted energy level to the second permitted level.
The value of

x

will be

Answer: 5

Solution

Energy of photon in transition from level

n_{i}

n_{f}

E = 13.6 (\frac{1}{n _{f}^{2}} - \frac{1}{n _{i}^{2}}) eV

For transition (i) from

n = 3

n = 2

E_{1} = 13.6 (\frac{1}{4} - \frac{1}{9}) = 13.6 \times \frac{5}{36}

For transition (ii) from

n = \infty

n = 2

E_{2} = 13.6 (\frac{1}{4} - 0) = 13.6 \times \frac{1}{4}

Ratio:

\frac{E _{1}}{E _{2}} = \frac{\frac{5}{36}}{\frac{1}{4}} = \frac{5}{36} \times 4 = \frac{5}{9}

Given

\frac{x}{x + 4} = \frac{5}{9}

9 x = 5 (x + 4)

9 x = 5 x + 20

4 x = 20

x = 5

JEE Main · 2022 · 25 Jul 2022 · Shift 2 · Integer answer

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JEE Main · 2022 · 25 Jul 2022 · Shift 2 · Modern Physics

Solution

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