If we consider an exited electron in state say n(of a hydrogen atom) which jumps to ground state, a photon is released .The wavelengthof photon is given by $λ^{-1}=R(1-1/n^2)$

According to law of conservation of momentum the momentum of atom and photon is conserved.

So $mv/n =h/λ +mv$ where v is speed of electron in ground state assuming photon is emitted in direction of motion of electron.

But the above equation is incorrect because λ is never negative.

Does it imply that atom is moving in direction opposite to direction in which photon is moving or photon can never be emitted in direction of motion of electron? Please explain in simple terms. I am a class 12 student.

• The concepts of electron and movement don't fit together. What do you want to express with the term $\frac{mv}{n}$? – Jasper Jun 3 '18 at 12:43
• The electron moves with speed v/n in nth allowed orbit. – Sai Saandeep Jun 3 '18 at 12:47
• Jasper sir. The electron moves. – Sai Saandeep Jun 3 '18 at 12:48
• @SaiSaandeep But in which direction? An S-state, for instance is spherical symmetric, which means the electron moves in all directions--not very useful for conservation of momentum questions. – JEB Jun 3 '18 at 16:30
• Related (but at requiring some stronger prerequisites): How does one account for the momentum of an absorbed photon?. – Emilio Pisanty Jun 3 '18 at 17:27

Assuming that the atom is initially at rest, according to the principle of conservation of linear momentum, we have

$$p_{atom} + p_{photon} = 0$$

The momentum $p_{atom}$ of the recoiling atom must be equal in magnitude and opposite in direction to the momentum of the emitted photon.

The momentum of emitted photon is $p_{photon} = h/λ$.

Hence, $p_{atom} = h/\lambda$ in the opposite direction.

Now, you can apply the equation for wavelength given by

$$\frac{1}{\lambda} = R (\frac{1}{n^2_1}- \frac{1}{n_2^2})$$ with $n_1<n_2$.

For a transition from higher state to ground state, you can simply substitute $n_1 = 1$ and $n_2 = n$ yielding $$p_{atom} = h R(1-\frac{1}{n^2})$$

You are correct to consider the recoil momentum of this reaction. The hydrogen atom gets a recoil momentum $m \vec v_r=-\hbar \vec k$. There is also a small recoil energy $E_r = \frac{1}{2} m v^2_r$, so the photon has a slightly smaller energy than the transition energy, $\hbar \omega = E_n-E_0-E_r$. Also angular momentum is conserved. Since the H ground state has $L=0$ and the photon has $S=1$, the H excited state must have $L=1$.

• What is $\vec k$ – Sai Saandeep Jun 3 '18 at 14:38
• It is the wave vector. It has the direction of the propagation and magnitude $2\pi / \lambda$. – my2cts Jun 3 '18 at 15:41