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(A) The energy difference between the levels $n=2$ and $n=1$ is given by $\Delta E = E_2 - E_1 = -3.4 \,eV - (-13.6 \,eV) = 10.2 \,eV$.By the principl

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$17$

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(A) The energy difference between the levels $n=2$ and $n=1$ is given by $\Delta E = E_2 - E_1 = -3.4 \,eV - (-13.6 \,eV) = 10.2 \,eV$.By the principle of conservation of momentum, the momentum of the emitted photon must be equal in magnitude to the recoil momentum of the hydrogen atom.$p_{	ext{atom}} = p_{	ext{photon}} = \frac{\Delta E}{c}$Since $p_{	ext{atom}} = m \cdot v$, where $m$ is the mass of the hydrogen atom and $v$ is the recoil speed:$v = \frac{\Delta E}{m \cdot c}$Substituting the given values:$v = \frac{10.2 \,eV}{(1.6 	imes 10^{-27} \,kg) 	imes (3 	imes 10^8 \,m/s)}$Convert $eV$ to Joules $(1 \,eV = 1.6 	imes 10^{-19} \,J)$:$v = \frac{10.2 	imes 1.6 	imes 10^{-19} \,J}{1.6 	imes 10^{-27} \,kg 	imes 3 	imes 10^8 \,m/s}$$v = \frac{10.2 	imes 10^{-19}}{3 	imes 10^{-19}} \,m/s = 3.4 \,m/s$Expressing $3.4 \,m/s$ as a fraction with denominator $5$:$v = \frac{3.4 	imes 5}{5} = \frac{17}{5} \,m/s$Comparing this with $\frac{x}{5} \,m/s$, we get $x = 17$.

When a hydrogen atom transitions from $n=2$ to $n=1$, it emits a photon. The recoil speed of the atom is $\frac{x}{5} \,m/s$. Find the value of $x$. (Use: mass of hydrogen atom $= 1.6 \times 10^{-27} \,kg$)

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