Solution

The initial momentum of the system is zero, so the momentum acquired by the nucleus must compensate for the momenta of the emitted particles. Since they move perpendicularly, we can set two coordinate axes along their directions of motion, then
$$p_x = p_e$$
$$p_y = p_{\nu} = \frac{E_{\nu}}{c} \quad \text{for a massless particle}$$
We can also write the invariant for the electron:
$$p^2 = \frac{E_e^2}{c^2} - m_e^2 c^2$$
And obtain the total momentum of the nucleus from the Pythagorean theorem:
$$p^2 = p_x^2 + p_y^2 = \frac{E_{\nu}^2}{c^2} + \frac{E_e^2}{c^2} - m_e^2 c^2$$

Now we write the expression for kinetic energy and the invariant for the nucleus.

$$E_K = E - M c^2$$

$$E^2 = c^2(p^2 + M^2 c^2) = E_{\nu}^2 + E_e^2 - m_e^2 c^4 + M^2 c^4$$
$$E_K = \sqrt{E_{\nu}^2 + E_e^2 - m_e^2 c^4 + M^2 c^4} - M c^2$$
Finally, recall that we are given the kinetic energy of the electron, not the total energy:
$$E_e = E_{eK} + m_e c^2$$
$$E_K = \sqrt{E_{\nu}^2 + (E_{eK} + m_e c^2)^2 - m_e^2 c^4 + M^2 c^4} - M c^2$$
Expand the parentheses
$$E_K = \sqrt{E_{\nu}^2 + E_{eK}(E_{eK} + 2 m_e c^2) + M^2 c^4} - M c^2$$
To find the numerical value, it is necessary to convert the rest energy into electron volts. For example, $M = 3.9 \cdot 10^{-22} \text{ g}\approx234 \ a.m.u.$- corresponds to uranium, $Mc^2\approx 3.19\cdot 10^5 \ MeV$
$$E_K\approx 13 \ eV$$

$Note:$

• There is a typo in Savchenko's answer, moreover, the author forgot about the numerical value.
• The author could not have known, but in 2015 the Nobel Prize in Physics was awarded to Takaaki Kajita and Arthur B. McDonald, who showed that the neutrino has a non-zero mass, albeit extremely small.