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(A) The energy of the photon emitted in the first line of the Lyman series ($n=2$ to $n=1$) is given by: $E = 13.6 \ eV 	imes (1 - 1/4) = 13.6 	imes

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$3.2$

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(A) The energy of the photon emitted in the first line of the Lyman series ($n=2$ to $n=1$) is given by: $E = 13.6 \ eV 	imes (1 - 1/4) = 13.6 	imes 0.75 = 10.2 \ eV$.Converting this to Joules: $E = 10.2 	imes 1.6 	imes 10^{-19} \ J = 1.632 	imes 10^{-18} \ J$.The momentum of the photon is $p = E/c = (1.632 	imes 10^{-18}) / (3 	imes 10^8) = 5.44 	imes 10^{-27} \ kg \ m/s$.By conservation of momentum,the recoil momentum of the hydrogen atom must be equal and opposite to the photon's momentum: $m_H v = p$.The mass of a hydrogen atom is $m_H \approx 1.67 	imes 10^{-27} \ kg$.Therefore,$v = p / m_H = (5.44 	imes 10^{-27}) / (1.67 	imes 10^{-27}) \approx 3.25 \ m/s$.

$1$. Burning candle	$i$. Line spectrum
$2$. Sodium vapour	$ii$. Continuous spectrum
$3$. Bunsen flame	$iii$. Band spectrum
$4$. Dark lines in solar spectrum	$iv$. Absorption spectrum

$A$ stationary hydrogen atom emits a photon corresponding to the first line of the Lyman series. The recoil velocity of the atom is nearly (in $m/s$)

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