A free hydrogen atom after absorbing a photon of wavelength gets excited from the state to the state . Immediately after that the electron jumps to state by emitting a photon of wavelength . Let the change in momentum of atom due to the absorption and

Question

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 be $Δ p_{a}$ and $Δ p_{e}$ , respectively. If $\frac{λ_{a}}{λ_{e}} = \frac{1}{5},$ which of the option(s) is/are correct

[Use $h c = 1242 eV nm$ ; $1 nm = 10^{- 9} m$ , $h$ and $c$ are Planck's constant and speed of light, respectively] [2019]

Smart Achievers · Accepted Answer

(2, 3)

(1) Change in linear momentum due to absorption,

$Δ P_{a} = \frac{h}{λ_{a}}$ $(∵ λ = \frac{h}{P})$

Change in linear momentum due to emission,

$Δ P_{e} = \frac{h}{λ_{e}}$ $∵$ $\frac{Δ P_{a}}{Δ P_{e}} = \frac{λ_{e}}{λ_{a}} = 5$ $(∵ \frac{λ_{a}}{λ_{e}} = \frac{1}{5} given)$

So, option (1) is wrong.

(2) Kinetic energy, $K_{n} \propto \frac{Z^{2}}{n^{2}}$ $∴$ $\frac{K_{2}}{K_{1}} = \frac{1^{2}}{2^{2}} = \frac{1}{4}$

So, (2) is correct.

(3) For absorption of energy, $n = 1$ to $n = 4$

$E_{4} - E_{1} = \frac{h c}{λ_{a}} = 13.6 (\frac{1}{1^{2}} - \frac{1}{4^{2}})$ $\dots (i)$

Per emission of energy, $n = m$ to $n = 4$

$E_{4} - E_{m} = \frac{h c}{λ_{e}} = 13.6 (\frac{1}{m^{2}} - \frac{1}{4^{2}})$ $\dots (i i)$

Dividing Eq. (ii) by Eq. (i),

$\frac{λ_{a}}{λ_{e}} = \frac{\frac{1}{m^{2}} - \frac{1}{16}}{1 - \frac{1}{16}} = \frac{1}{5}$

$\Rightarrow \frac{1}{m^{2}} - \frac{1}{16} = \frac{1}{5} \times \frac{15}{16}$ $\Rightarrow \frac{1}{m^{2}} = \frac{3}{16} + \frac{1}{16} = \frac{4}{16}$

$∴$ $m = 2$

So, (3) is correct.

(4) Now from Eq. (ii),

$\frac{h c}{λ_{e}} = 13.6 (\frac{1}{2^{2}} - \frac{1}{4^{2}}) = 13.6 (\frac{1}{4} - \frac{1}{16})$

$∴$ $λ_{e} = \frac{1242 \times 16}{13.6 \times 3} \approx 486.2 nm$

So, option (4) is wrong.

Solution

1 $Δ p_{a} / Δ p_{e} = \frac{1}{2}$

2 The ratio of kinetic energy of the electron in the state $n = m$ to the state $n = 1$ is $\frac{1}{4}$

3 $m = 2$

4 $λ_{e} = 418 nm$

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Solution

1 Δpa/Δpe=12

2 The ratio of kinetic energy of the electron in the state n=m to the state n=1 is 14

3 m=2

4 λe=418 nm

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1 $Δ p_{a} / Δ p_{e} = \frac{1}{2}$

2 The ratio of kinetic energy of the electron in the state $n = m$ to the state $n = 1$ is $\frac{1}{4}$

3 $m = 2$

4 $λ_{e} = 418 nm$