JEE Advanced · Physics · 27. Atomic Physics

A free hydrogen atom after absorbing a photon of wavelength $λ_{a}$ gets excited from the state $n = 1$ to the state $n = 4 .$ Immediately after that the electron jumps to $n = m$ state by emitting a photon of wavelength $λ_{e} .$ Let the change in momentum of atom due to the absorption and the emission are $Δ p_{a}$ and $Δ p_{e},$ respectively. If $\frac{λ_{a}}{λ_{e}} = \frac{1}{5} .$ Which of the option(s) is/are correct?
[Use $h c = 1242 e V n m, 1 n m = 10^{- 9} m,$ $h$ and $c$ are Planck's constant and speed of light, respectively]

A $λ_{e} = 418 n m$
B The ratio of kinetic energy of the electron in the state $n = m$ to the state $n = 1$ is $\frac{1}{4}$
C $m = 2$
D $\frac{Δ p_{a}}{Δ p_{e}} = \frac{1}{2}$

Verified Solution

Answer & Solution

Correct Answer

(C) $m = 2$

Step-by-step Solution

Detailed explanation

Energy for transition of electron from one orbit to other is given by-

E_{2} - E_{1} = (13.6) Z^{2} (\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}) = \frac{h c}{λ}

\Rightarrow \frac{1}{λ} = \frac{13.6}{h c} . Z^{2} (\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}})

Now as per question

\frac{1}{λ_{a}} = \frac{13.6}{h c} . Z^{2} (\frac{1}{1^{2}} - \frac{1}{4^{2}})

and

\frac{1}{λ_{e}} = \frac{13.6}{h c} . Z^{2} (\frac{1}{m^{2}} - \frac{1}{4^{2}})

\Rightarrow \frac{λ_{e}}{λ_{a}} = \frac{(1 - \frac{1}{16})}{\frac{1}{m^{2}} - \frac{1}{16}} = \frac{(15) m^{2}}{16 - m^{2}}

But

\frac{λ_{a}}{λ_{e}} = \frac{1}{5} (g i v e n)

\Rightarrow \frac{16 - m^{2}}{15 m^{2}} = \frac{1}{5} \Rightarrow 16 - m^{2} = 3 m^{2}

\Rightarrow m^{2} = 4

\Rightarrow m = 2 \Rightarrow (C)

is correct
Now,

\frac{1}{λ_{e}} = \frac{13.6}{h c} . Z^{2} (\frac{1}{m^{2}} - \frac{1}{16})

$=\frac{13.6}{1242} \times 1\left(\frac{1}{4}-\frac{1}{16}\right) \quad($ as $h c=1242$ and $z=1)$

\Rightarrow λ_{e} = \frac{1242}{13.6} \times \frac{16}{3} = 487 n m \Rightarrow (A)

is incorrect
Now, kinetic energy of electron

K \propto \frac{1}{n^{2}}

\Rightarrow \frac{K_{2}}{K_{1}} = \frac{1^{2}}{2^{2}} = \frac{1}{4} \Rightarrow (B)

is correct
Also,

∆ P = \frac{h}{λ}

\Rightarrow \frac{∆ P_{a}}{∆ P_{e}} = \frac{λ_{e}}{λ_{a}} = 5

\Rightarrow (D)

is incorrect.

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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5.00 mL of 0.10 M oxalic acid solution taken in a conical flask is titrated against NaOH from a burette using phenolphthalein indicator. The volume of NaOH required for the appearance of permanent faint pink color is tabulated below for five experiments. What is the concentration, in molarity, of the NaOH solution?
Exp. No. Vol. of NaOH (mL)
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2 10.5
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Exp. No.	Vol. of NaOH (mL)
1	12.5
2	10.5
3	9.0
4	9.0
5	9.0