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Question: The recoil speed of hydrogen atom after it emits a photon in going from \(n = 2\) state to \(n = 1\)...

The recoil speed of hydrogen atom after it emits a photon in going from n=2n = 2 state to n=1n = 1 state is nearly [TakeR∞=1.1×107m−1{R_\infty } = 1.1 \times {10^7}{m^{ - 1}} and h=6.63×10−34Jsh = 6.63 \times {10^{ - 34}}Js]

A.)1.5ms−11.5m{s^{ - 1}}
B.)3.3ms−13.3m{s^{ - 1}}
C.)4.5ms−14.5m{s^{ - 1}}
D.)6.6ms−16.6m{s^{ - 1}}

Explanation

Solution

Hint – Start the solution by explaining why the photon is released when an electron is released and when an electron jumps to a more stable orbit. Then use Rydberg equation1λ=R⋅(1nf2−1ni2)\dfrac{1}{\lambda } = R \cdot \left( {\dfrac{1}{{n_f^2}} - \dfrac{1}{{n_i^2}}} \right), followed by Einstein Planck relation E=h⋅cλE = h \cdot \dfrac{c}{\lambda } and at last compare momentum of the photon with that of the hydrogen atom.

Complete step by step answer:
Using the Rydberg Equation, we get
1λ=R⋅(1nf2−1ni2)\dfrac{1}{\lambda } = R \cdot \left( {\dfrac{1}{{n_f^2}} - \dfrac{1}{{n_i^2}}} \right)(Equation 1)
Here λ=\lambda = wavelength of emitted photon,

RR is the Rydberg constant=1.1×107m−1 = 1.1 \times {10^7}{m^{ - 1}},

nf{n_f} is the number of final orbit
ni{n_i} is the number of initial orbit

Substituting these values in equation 1, we get,

1λ=1.1×107⋅(112−122)\dfrac{1}{\lambda } = 1.1 \times {10^7} \cdot \left( {\dfrac{1}{{{1^2}}} - \dfrac{1}{{{2^2}}}} \right)

⇒1λ=3.3×1074 \Rightarrow \dfrac{1}{\lambda } = \dfrac{{3.3 \times {{10}^7}}}{4}(Equation 2)

Now, using Einstein Planck relation, we have,
E=h⋅cλE = h \cdot \dfrac{c}{\lambda }(Equation 3)

Where

E=E = Energy of photon,
c=c = Speed of light 3⋅108ms−13 \cdot {10^8}m{s^{ - 1}}

Substituting value of 1λ\dfrac{1}{\lambda }from equation 2 in equation 3, we get

E=6.63×10−34×3×108×3.3×1074E = 6.63 \times {10^{ - 34}} \times 3 \times {10^8} \times \dfrac{{3.3 \times {{10}^7}}}{4}
⇒E=16.4×10−19J\Rightarrow E = 16.4 \times {10^{ - 19}}J
Mass of Hydrogen(mH)({m_H}) =1.67×10−27Kg = 1.67 \times {10^{ - 27}}Kg

Let velocity of hydrogen atom after ejection of photon bevH{v_H}

We know that momentum of photon=momentum of hydrogen atom=Ec = \dfrac{E}{c}

So, mH⋅vH=Ec{m_H} \cdot {v_H} = \dfrac{E}{c}

⇒vH=Ec⋅mH=16.4×10−193×108×1.67×10−27 \Rightarrow {v_H} = \dfrac{E}{{c \cdot {m_{^H}}}} = \dfrac{{16.4 \times {{10}^{ - 19}}}}{{3 \times {{10}^8} \times 1.67 \times {{10}^{ - 27}}}}
⇒vH=3.3ms−1\Rightarrow {v_H} = 3.3m{s^{ - 1}}

Hence, option B is the correct choice.

Additional Information: Photon – They are small packets that are the basic components of any electromagnetic wave. They carry energy and are very important when we study the particle nature of light. Photons are always in motion and they have no mass.
The concept of photons is very essential to understand the problem given above. Remember, every system wants to attain the least amount of energy possible. When an electron moves from an excited state to a less excited state or to a stable state, it releases excess energy in the form of photons that are ejected out of the atom.
The next concept we will use is the concept of conservation of energy, which implies that energy can neither be created nor destroyed and can be only changed from one form to another.

Note: We discussed how every entity in the universe wants to have the least possible energy, and we used the concept of conservation of energy to figure out the solution. In such problems we always consider ideal conversion of energy, but in reality some heat is also produced. This gives birth to the idea that the total heat energy in the universe is always increasing.