Consider a helium (He) atom that absorbs a ph | vedclass

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Question

(A) The momentum of the photon is given by $p = \frac{h}{\lambda}$.Substituting the values: $p = \frac{6.6 	imes 10^{-34} \ J \ s}{330 	imes 10^{-9}

Vedclass · Accepted Answer

$30$

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(A) The momentum of the photon is given by $p = \frac{h}{\lambda}$.Substituting the values: $p = \frac{6.6 	imes 10^{-34} \ J \ s}{330 	imes 10^{-9} \ m} = 2 	imes 10^{-27} \ kg \ m \ s^{-1}$.According to the law of conservation of momentum,the momentum gained by the $He$ atom is equal to the momentum of the absorbed photon.$p_{atom} = m 	imes v$,where $m$ is the mass of a single $He$ atom.$m = \frac{	ext{Molar mass}}{	ext{Avogadro number}} = \frac{4 	imes 10^{-3} \ kg \ mol^{-1}}{6 	imes 10^{23} \ mol^{-1}} = \frac{2}{3} 	imes 10^{-26} \ kg$.Equating the momenta: $2 	imes 10^{-27} = (\frac{2}{3} 	imes 10^{-26}) 	imes v$.$v = \frac{2 	imes 10^{-27} 	imes 3}{2 	imes 10^{-26}} = 0.3 \ m \ s^{-1}$.Converting to $cm \ s^{-1}$: $0.3 \ m \ s^{-1} = 30 \ cm \ s^{-1}$.

Metal	$Li, Na, K, Mg, Cu, Ag, Fe, Pt, W$
$\phi \ (eV)$	$2.4, 2.3, 2.2, 3.7, 4.8, 4.3, 4.7, 6.3, 4.75$

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