Two hydrogen atoms are in excited state with electrons in`n=2` state.First one is moving to wards left and emits a photon.' of energy `E_(1)` towards right. Second one is moving towards right with same speed and emits a photon of energy `E_(2)` towards right. Taking recoil of nucleus.into account during_emission process :

Options

A. `E_(1) gt E_(2)`
B. `E_(1) lt E_(2)`
C. `E_(1) = E_(2)`
D. information insufficient

Explanation

To solve the problem, we will analyze the situation step by step, considering the conservation of momentum and energy during the emission of photons by the two hydrogen atoms.

### Step 1: Understand the Initial Conditions
We have two hydrogen atoms, both in an excited state with their electrons in the \( n = 2 \) state. One atom is moving to the left and emits a photon \( E_1 \) towards the right, while the other atom is moving to the right and emits a photon \( E_2 \) also towards the right.

**Hint:** Identify the initial states and directions of motion of the atoms and the emitted photons.

### Step 2: Apply Conservation of Momentum
When the first hydrogen atom emits a photon \( E_1 \), the conservation of momentum must hold. Let the mass of the hydrogen atom be \( m \) and its initial velocity be \( v \). After emitting the photon, let the new velocity of the first atom be \( v' \).

The momentum before emission is:
\[ mv \]

The momentum after emission is:
\[ mv' + \frac{E_1}{c} \]
where \( \frac{E_1}{c} \) is the momentum of the emitted photon (using \( p = \frac{E}{c} \)).

Setting these equal gives:
\[ mv = mv' + \frac{E_1}{c} \]

**Hint:** Write down the equations for momentum conservation before and after the emission.

### Step 3: Analyze the Second Atom
For the second hydrogen atom moving to the right and emitting a photon \( E_2 \), we apply the same conservation of momentum principle. Let its new velocity after emission be \( v'' \).

The momentum before emission is:
\[ mv \]

The momentum after emission is:
\[ mv'' + \frac{E_2}{c} \]

Setting these equal gives:
\[ mv = mv'' + \frac{E_2}{c} \]

**Hint:** Set up a similar equation for the second atom as you did for the first.

### Step 4: Relate the Velocities
From the momentum equations, we can express \( v' \) and \( v'' \):
1. For the first atom:
\[ v' = v - \frac{E_1}{mc} \]
2. For the second atom:
\[ v'' = v - \frac{E_2}{mc} \]

**Hint:** Isolate the new velocities in terms of the emitted photon energies.

### Step 5: Compare the Velocities
Since both atoms are initially moving with the same speed \( v \), we can compare \( v' \) and \( v'' \):
\[ v' < v \quad \text{and} \quad v'' < v \]

Since \( v' \) and \( v'' \) depend on the energies of the emitted photons, we can conclude:
\[ v' = v - \frac{E_1}{mc} \quad \text{and} \quad v'' = v - \frac{E_2}{mc} \]

**Hint:** Consider how the emitted energy affects the final velocities.

### Step 6: Apply Conservation of Energy
Next, we apply conservation of energy. The total energy before emission is equal to the total energy after emission:
1. For the first atom:
\[ \frac{1}{2} mv^2 = \frac{1}{2} mv'^2 + E_1 \]
2. For the second atom:
\[ \frac{1}{2} mv^2 = \frac{1}{2} mv''^2 + E_2 \]

**Hint:** Write down the equations for energy conservation for both atoms.

### Step 7: Compare the Energies
From the energy equations, we can see that since \( v' < v'' \), it implies:
\[ E_1 < E_2 \]

Thus, the energy of the photon emitted by the first atom is less than that emitted by the second atom.

**Hint:** Use the relationship between the velocities and energies to conclude which photon has more energy.

### Conclusion
The final conclusion is that the energy of the photon emitted by the first hydrogen atom \( E_1 \) is less than the energy of the photon emitted by the second hydrogen atom \( E_2 \).

**Final Answer:** \( E_1 < E_2 \)