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Step-by-Step Solution
Step 1: Determine the Energy of the Emitted Photon
In a hydrogen atom, the energy of an electron in the n^\mathrm{th} level is given by
E_n = -\frac{13.6\,\text{eV}}{n^2} .
For the transition from n = 2 to n = 1 , the energy difference (the emitted photon’s energy) is:
\Delta E = E_1 - E_2
= \left(-\frac{13.6\,\text{eV}}{1^2}\right)
- \left(-\frac{13.6\,\text{eV}}{2^2}\right).
Substituting values,
E_1 = -13.6\,\text{eV},
\quad
E_2 = -3.4\,\text{eV}.
Therefore,
\Delta E = -13.6\,\text{eV} - \bigl(-3.4\,\text{eV}\bigr)
= -13.6\,\text{eV} + 3.4\,\text{eV}
= -10.2\,\text{eV}.
The negative sign indicates emission, so the photon’s energy is 10.2\,\text{eV} .
Step 2: Convert Photon Energy from eV to Joules
Use the conversion 1\,\text{eV} = 1.602 \times 10^{-19}\,\text{J} . Thus,
E_{\text{photon}}
= 10.2\,\text{eV} \times 1.602 \times 10^{-19}\,\frac{\text{J}}{\text{eV}}
= 1.632 \times 10^{-18}\,\text{J}.
Step 3: Calculate the Photon’s Momentum
The momentum of a photon is given by
p = \frac{E}{c} , where c is the speed of light ( \approx 3 \times 10^8\,\text{m/s} ). Hence,
p_{\text{photon}}
= \frac{1.632 \times 10^{-18}\,\text{J}}{3 \times 10^8\,\text{m/s}}
= 5.44 \times 10^{-27}\,\text{kg m/s}.
Step 4: Relate Photon Momentum to Hydrogen Atom Recoil
By conservation of momentum, the recoiling hydrogen atom (mass m ) has momentum
p_{\text{atom}} = p_{\text{photon}} . Thus,
m \cdot v = p_{\text{photon}},
where v is the recoil speed and m is the mass of the hydrogen atom
( 1.6 \times 10^{-27}\,\text{kg} ). Therefore,
v
= \frac{p_{\text{photon}}}{m}
= \frac{5.44 \times 10^{-27}\,\text{kg m/s}}{1.6 \times 10^{-27}\,\text{kg}}
\approx 3.4\,\text{m/s}.
Step 5: Identify the Value of x
The recoil speed is given in the form \frac{x}{5}\,\text{m/s} . We obtained
v \approx 3.4\,\text{m/s} . Hence,
\frac{x}{5} = 3.4
\quad \Longrightarrow \quad
x = 3.4 \times 5 = 17.
Final Answer
x = 17 .
