
An electron of a stationary hydrogen atom passes from the fifth energy level to the ground level. The velocity that the atom of mass $m$ acquired as a result of photon emission will be:
($R$ is Rydberg constant and $h$ is Planck's constant)
(A) $\dfrac{{25m}}{{24hR}}$
(B) $\dfrac{{24m}}{{25hR}}$
(C) $\dfrac{{24hR}}{{25m}}$
(D) $\dfrac{{25hR}}{{24m}}$
Answer
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Hint The electrons revolve the atom in a specific orbit which is also known as energy levels. The first energy level is called the ground state, then comes the first excited state, then the second excited state, and so on. The transition of electrons from a higher energy state to a lower energy state results in losing energy while the transition from a lower energy state to a higher energy state results in gaining energy.
Complete Step by step answer
We need to remember the formula correlating the wavelength of an emitted photon with the orbital levels of the electron transition. This is given by
$\dfrac{1}{\lambda } = R(\dfrac{1}{{{n_1}^2}} - \dfrac{1}{{{n_2}^2}})$
where $R$ is Rydberg constant,
${n_1}$ and ${n_2}$ are the final and initial orbitals respectively and
$\lambda $is the wavelength of the emitted photon.
It is given in the question that the electron drops from the fifth to the ground state of the atom.
Thus, substituting ${n_1} = 1$ and ${n_2} = 5$, in the above equation, we get
$\dfrac{1}{\lambda } = R(\dfrac{1}{{{1^2}}} - \dfrac{1}{{{5^2}}}) = \dfrac{{24R}}{{25}}$.
We will use this value of
$\dfrac{1}{\lambda } = \dfrac{{24R}}{{25}}$
in the equation $E = \dfrac{{hc}}{\lambda }$, so that we get
$E = \dfrac{{24R}}{{25}}hc$.
Since the momentum of the atom is given by
$p = \dfrac{E}{c}$,
we get $p = \dfrac{E}{c} = \dfrac{{24R}}{{25}}h$.
The momentum of the atom after the photon emission has been found as above, given by the equation $p = \dfrac{{24R}}{{25}}h$.
Since we require the velocity of the atom, we will divide the momentum with the mass of the atom which is given to be $m$.
$\dfrac{p}{m} = \dfrac{{24Rh}}{{25m}}$.
Therefore the correct answer is option (C) $\dfrac{{24hR}}{{25m}}$.
Note Remember that in case of problems regarding subatomic particles, we should always consider the rest mass of the particle, unless mentioned otherwise. Hence the values of rest masses of electron, proton, and neutron should always be memorized. For this sum, the rest mass of an electron is $9.1 \times {10^{ - 31}}kg$.
Complete Step by step answer
We need to remember the formula correlating the wavelength of an emitted photon with the orbital levels of the electron transition. This is given by
$\dfrac{1}{\lambda } = R(\dfrac{1}{{{n_1}^2}} - \dfrac{1}{{{n_2}^2}})$
where $R$ is Rydberg constant,
${n_1}$ and ${n_2}$ are the final and initial orbitals respectively and
$\lambda $is the wavelength of the emitted photon.
It is given in the question that the electron drops from the fifth to the ground state of the atom.
Thus, substituting ${n_1} = 1$ and ${n_2} = 5$, in the above equation, we get
$\dfrac{1}{\lambda } = R(\dfrac{1}{{{1^2}}} - \dfrac{1}{{{5^2}}}) = \dfrac{{24R}}{{25}}$.
We will use this value of
$\dfrac{1}{\lambda } = \dfrac{{24R}}{{25}}$
in the equation $E = \dfrac{{hc}}{\lambda }$, so that we get
$E = \dfrac{{24R}}{{25}}hc$.
Since the momentum of the atom is given by
$p = \dfrac{E}{c}$,
we get $p = \dfrac{E}{c} = \dfrac{{24R}}{{25}}h$.
The momentum of the atom after the photon emission has been found as above, given by the equation $p = \dfrac{{24R}}{{25}}h$.
Since we require the velocity of the atom, we will divide the momentum with the mass of the atom which is given to be $m$.
$\dfrac{p}{m} = \dfrac{{24Rh}}{{25m}}$.
Therefore the correct answer is option (C) $\dfrac{{24hR}}{{25m}}$.
Note Remember that in case of problems regarding subatomic particles, we should always consider the rest mass of the particle, unless mentioned otherwise. Hence the values of rest masses of electron, proton, and neutron should always be memorized. For this sum, the rest mass of an electron is $9.1 \times {10^{ - 31}}kg$.
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