JEE Main · 2019 · 12 Apr 2019 · Shift 2 · Modern Physics

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The electron in a hydrogen atom first jumps from the third excited state to the second excited state and subsequently to the first excited state. The ratio of the respective wavelengths,

λ_{1} / λ_{2}

, of the photons emitted in this process is:

(A) $20/7$ ✓ Correct
(B) $7/5$
(C) $9/7$
(D) $27/5$

Answer: (A)

Solution

The third excited state is

n = 4

, the second excited is

n = 3

, the first excited is

n = 2

. For the first jump

4 \to 3

\frac{1}{λ _{1}} = R (\frac{1}{9} - \frac{1}{16}) = R \frac{7}{144}

, and for

3 \to 2

\frac{1}{λ _{2}} = R (\frac{1}{4} - \frac{1}{9}) = R \frac{5}{36}

. Dividing,

\frac{λ _{1}}{λ _{2}} = \frac{5/36}{7/144} = \frac{5 \times 144}{36 \times 7} = \frac{20}{7}

JEE Main · 2019 · 12 Apr 2019 · Shift 2 · Single correct option

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