Solution

Center-of-mass energy in the symmetric collider

Each electron has total energy $E = N m_e c^2$.
When two identical particles collide head‑on with opposite momenta, the total energy in the center‑of‑mass (CM) system is simply the sum of their total energies:

$E_{\text{CM}} = 2E = 2N m_e c^2$

This is the energy available for the interaction.

Center-of-mass energy in the fixed target

Now we have an incident electron with total energy$E_{\text{lab}}$ and a target electron at rest (energy $m_e c^2)$
The square of the center‑of‑mass energy for this system is calculated using the relativistic invariant s

$s = (E_{\text{lab}} + m_e c^2)^2 - (p_{\text{lab}} c)^2$

where$p_{\text{lab}}$is the momentum of the incident electron.

Using the relation
$E_{\text{lab}}^2 - (p_{\text{lab}} c)^2 = (m_e c^2)^2$
we expand:

$s = E_{\text{lab}}^2 + 2E_{\text{lab}}m_e c^2 + (m_e c^2)^2 - (p_{\text{lab}} c)^2 = 2(m_e c^2)^2 + 2E_{\text{lab}}m_e c^2$

The center‑of‑mass energy is E_{\text{CM}} = \sqrt{s}.

For high energies $(E_{\text{lab}} \gg m_e c^2)$this approximates to:

$E_{\text{CM}} \approx \sqrt{2m_e c^2 E_{\text{lab}}}$

Equating the center‑of‑mass energies

We want the available energy in the fixed target to be the same as in the symmetric collider:

$\sqrt{2m_e c^2 E_{\text{lab}}} = 2N m_e c^2$

Squaring both sides:

$2m_e c^2 E_{\text{lab}} = 4 N^2 (m_e c^2)^2 \quad\Rightarrow\quad E_{\text{lab}} = 2 N^2 m_e c^2$

The total energy that the incident electron must have is $E_{\text{lab}} = 2 N^2 m_e c^2$
Since N = 1000, we get

$E_{\text{lab}} = 2 \times (1000)^2\, m_e c^2 = 2 \times 10^6\, m_e c^2$

That is, the electron energy must be 2 million times its rest energy.

Comparing with the energy of each electron in the collider $(N m_e c^2 = 1000\, m_e c^2)$, the increase factor is:

$\frac{E_{\text{lab}}}{N m_e c^2} = \frac{2 N^2}{N} = 2N = 2000$

Therefore, an energy 2000 times greater than that of each electron in the colliding beams is needed.