A hydrogen atom changes its state from . Due to recoil, the percentage change in the wavelength of emitted light is approximately . The value of nnn is

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Solution

Q.
A hydrogen atom changes its state from $n = 3 t o n = 2$ . Due to recoil, the percentage change in the wavelength of emitted light is approximately $1 \times 10^{- n}$ . The value of n is ____.

[Given $R h c = 13.6 eV, h c = 1242 eVnm, h = 6.6 \times 10^{- 34}$ mass of the hydrogen atom = $1.6 \times 10^{- 27} kg$ ] [2024]

Ans.
(7)

For $n = 3$ to $n = 2$

$Δ E = 13.6 (\frac{1}{2^{2}} - \frac{1}{3^{2}}) = 1.9 eV$

$Δ E = \frac{h c}{λ} \Rightarrow λ = \frac{h c}{Δ E}$

If recoil takes place

$P_{i} = P_{f} \Rightarrow 0 = - m v + \frac{h}{λ'} \Rightarrow v = \frac{h}{m λ'}$

$Δ E = \frac{1}{2} m v^{2} + \frac{h c}{λ'} = \frac{1}{2} m {(\frac{h}{m λ'})}^{2} + \frac{h c}{λ'}$

Now, $Δ E = \frac{h^{2}}{2 m λ'^{2}} + \frac{h c}{λ'}$

$λ'^{2} Δ E - h c λ' - \frac{h^{2}}{2 m} = 0$

$λ' = \frac{h c \pm \sqrt{h^{2} c^{2} + \frac{4 Δ E h^{2}}{2 m}}}{2 Δ E}$

$λ' = \frac{h c \pm h c \sqrt{1 + \frac{2 Δ E}{m c^{2}}}}{2 Δ E}$

$\frac{λ'}{λ} = \frac{1 + {(1 + \frac{2 Δ E}{m c^{2}})}^{\frac{1}{2}}}{2} = \frac{1 + 1 + \frac{Δ E}{m c^{2}}}{2}$

$\frac{λ'}{λ} = 1 + \frac{Δ E}{2 m c^{2}}$

$\frac{λ' - λ}{λ} = \frac{Δ E}{2 m c^{2}} = \frac{1.9 \times 1.6 \times 10^{- 19}}{2 \times 1.67 \times 10^{- 27} \times 9 \times 10^{16}} = 10^{- 9}$

$∴$ $% change \approx 10^{- 7}$

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