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(C) The kinetic energy $K$ of the emitted electron is given by the de Broglie wavelength formula: $\lambda = \frac{h}{\sqrt{2mK}}$.Rearranging for $K$

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(C) The kinetic energy $K$ of the emitted electron is given by the de Broglie wavelength formula: $\lambda = \frac{h}{\sqrt{2mK}}$.Rearranging for $K$: $K = \frac{h^2}{2m\lambda^2}$.Substituting the given values: $K = \frac{(6.626 	imes 10^{-34})^2}{2 	imes 9.1 	imes 10^{-31} 	imes (3.3 	imes 10^{-10})^2} \approx 2.215 	imes 10^{-18} \ J$.The minimum energy required to escape from the ground state of $H$ atom is the ionization energy: $E_{req} = 13.6 \ eV = 13.6 	imes 1.602 	imes 10^{-19} \ J \approx 2.179 	imes 10^{-18} \ J$.The total energy absorbed is $E_{abs} = E_{req} + K$.The ratio is $\frac{E_{abs}}{E_{req}} = 1 + \frac{K}{E_{req}} = 1 + \frac{2.215 	imes 10^{-18}}{2.179 	imes 10^{-18}} \approx 1 + 1.016 = 2.016$.Rounding to the nearest integer,the value is $2$.

List-$I$	List-$II$
$A$. Nodes	$I$. Three dimensional shape of the orbital
$B$. Subsidiary quantum number	$II$. Significant only for motion of microscopic objects
$C$. White light	$III$. $\|\psi\|^2$ is zero
$D$. Heisenberg uncertainty principle	$IV$. Spin state of electron
	$V$. Continuous spectrum

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