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Step-by-Step Solution
Step 1: Determine the wavelength of the emitted photon
For a hydrogen atom (with atomic number Z = 1), the Rydberg formula for the inverse of wavelength is:
$ \frac{1}{\lambda} = R Z^2 \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right). $
Here, the electron transitions from the fifth energy level ($n_2 = 5$) to the ground level ($n_1 = 1$):
$ \frac{1}{\lambda} = R \left(\frac{1}{1^2} - \frac{1}{5^2} \right) = R \left(1 - \frac{1}{25}\right) = R \left(\frac{25}{25} - \frac{1}{25}\right) = \frac{24R}{25}. $
Step 2: Use conservation of linear momentum
When the photon is emitted, the total linear momentum of the system must remain conserved:
Momentum of photon = Momentum of atom.
The momentum of a photon is given by $ \frac{h}{\lambda} $, where $h$ is Planck's constant and $\lambda$ is the wavelength of the photon. Let $m$ be the mass of the hydrogen atom and $v$ its recoil velocity:
$ \frac{h}{\lambda} = m v. $
Step 3: Solve for the velocity of the recoiling atom
Rearrange the above equation to find $v$:
$ v = \frac{h}{m\,\lambda}. $
Substitute $ \frac{1}{\lambda} = \frac{24R}{25} $:
$ \lambda = \frac{1}{\frac{24R}{25}} = \frac{25}{24R}. $
Hence,
$ v = \frac{h}{m} \times \frac{24R}{25} = \frac{24\,h\,R}{25\,m}. $
Therefore, the velocity of the hydrogen atom after the emission is:
$ \boxed{ \frac{24\,h\,R}{25\,m} }. $
