Answer
$6.45\times 10^{-12 }\;\rm m$
Work Step by Step
We know that the kinetic energy of the electron due to accelerating by 35 000 V is given by
$$KE=-q\Delta V=-e\Delta V$$
Plugging the known;
$$KE=-\left(-1.6\times10^{-19}\right)\times 35 0000$$
$$KE=\bf 5.6\times10^{-15}\;\rm J=\bf 35000\;\rm eV\tag 1$$
We know that the energy of the electron is given by
$$E=KE+m_ec^2=\sqrt{p^2c^2+m_e^2c^4}$$
Thus,
$$ \left(KE^2+m_ec^2\right)^2=p^2c^2+m_e^2c^4$$
$$ KE^2+2KEm_ec^2+\color{red}{m_e^2c^4 }=p^2c^2+\color{red}{m_e^2c^4 }$$
$$ KE^2+2KEm_ec^2 =p^2c^2 $$
Therefore, its momentum is given by
$$p=\sqrt{\dfrac{ KE^2+2KEm_ec^2}{c^2 }}\tag 2$$
We know that the electron's momentum is given by
$$p=\dfrac{h}{\lambda}$$
So, its wavelength is given by
$$\lambda=\dfrac{h}{p}$$
Plugging from (2);
$$\lambda=\dfrac{h}{\sqrt{\dfrac{ KE^2+2KEm_ec^2}{c^2 }}}=\dfrac{h}{\dfrac{\sqrt{ KE^2+2KEm_ec^2}}{c }}$$
$$\lambda=\dfrac{hc}{\sqrt{ KE^2+2KEm_ec^2}} $$
Plugging the known;
$$\lambda=\dfrac{6.626\times10^{-34}\times3\times10^8}{\sqrt{ ( 5.6\times10^{-15})^2+\left[2\times 5.6\times10^{-15}\times 9.11\times10^{-31}\times (3\times10^8)^2\right]}} $$
$$\lambda=\color{red}{\bf 6.45\times 10^{-12}}\;\rm m$$
Now since the wavelength of the electron is less than the diameter of the neck of the tube $(\lambda\lt \lt D)$, then the diffraction effect is negligible.
And since the kinetic energy of the electron is greater than its energy at rest, so it is relativistic.
