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If the wavelength for an electron emitted from H-atom is $3.3 \times 10^{- 10} m,$ then energy absorbed by the electron in its ground state compared to minimum energy required for it’s escape from the atom, is $\frac{x}{5}$ times what is $x$ (Nearest integer) {Given, $h = 6.626 \times 10^{- 34} J s$ } [Mass of electron $= 9.1 \times 10^{31} k g$ ]

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answer is 4.

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Detailed Solution

$λ = \frac{h}{\sqrt{2 m k}}$

$\Rightarrow k = \frac{h^{2}}{2 m λ^{2}}$

$= \frac{{(6.626 \times 10^{- 34})}^{2}}{2 \times 9.1 \times 10^{- 31} \times {(3.3 \times 10^{- 10})}^{2}}$

$= \frac{43.90 \times 10^{- 68}}{198.198 \times 10^{- 51}}$

$= 0.221 \times 10^{- 17}$

$= 2.21 \times 10^{- 18}$

$E_{a b s} = E_{r e q} + K$

$\frac{E_{a b s}}{E_{r e q}} = 1 + \frac{k}{E_{r e q}} = 1 + \frac{2.215 + 10^{- 18}}{13.6 \times 1.602 \times 10^{19}}$

$= 20.01 ≃ \frac{20}{5} = 4$

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