a) In order to solve this problem we are going to use the de Broglie relation which says that

$$
lambda=frac{h}{p}=frac{h}{mv}
$$

$$
lambda=frac{6.63times 10^{-34}}{7times 8.5}
$$

Which gives that

$$
boxed{lambda=1.11times 10^{-35}textrm{ m}}
$$

b) The effects are not observable since the de Broglie wavelength is too small.

$$
textrm{a) }lambda=1.11times 10^{-35}textrm{ m}
$$

$$
textrm{b) The de Broglie wavelength is too small to be observable. }
$$

In order to solve this problem we are going to use the conservation of energy to find the speed of the electron first and then we are going to use de Broglie relation to find the de Broglie wavelength.

$$
frac{mv^2}{2}=qV
$$

which makes that the speed is given as

$$
v=sqrt{frac{2qV}{m}}=sqrt{frac{2times 1.6times 10^{-19}times 250}{9.11times 10^{-31}}}
$$

Finally,

$$
boxed{v=9.4times 10^6, frac{textrm{m}}{textrm{s}}}
$$

Now, de Broglie’s wavelength is given by

$$
lambda=frac{h}{mv}=frac{6.63times 10^{-34}}{9.11times 10^{-31}times 9.4times 10^6}
$$

which gives that

$$
boxed{lambda=0.077times 10^{-9}textrm{ m}}
$$

$$
v=9.4times 10^6, frac{textrm{m}}{textrm{s}}
$$

$$
lambda=0.077times 10^{-9}textrm{ m}
$$

In order to solve this problem we are going to use the definition of the de Broglie ‘s wavelength to find the corresponding velocity which in turn will tell us what is required kinetic energy i.e. potential energy. Let’s do it

$$
lambda=frac{h}{mv}
$$

So the velocity is given as

$$
v=frac{h}{mlambda}
$$

Now, we can insert this into the expression for kinetic energy

$$
K=frac{mv^2}{2}=frac{mh^2}{2m^2lambda^2}=frac{h^2}{2mlambda^2}
$$

Now, we can plug in the values to have that

$$
K=frac{h^2}{2mlambda^2}=frac{6.63^2times 10^{-68}}{2times 9.11times 10^{-31}times 0.125^2times 10^{-18}}
$$

Which gives us that

$$
K=154times 10^{-19}textrm{ J}
$$

Now, the voltage needed for an electron to achieve this kinetic energy is

$$
V=frac{K}{e}=frac{154times 10^{-19}}{1.6times 10^{-19}}
$$

Finally, we have that

$$
boxed{V=96.5 textrm{ V}}
$$

$$
V=96.5 textrm{ V}
$$

In order to solve this problem we are going to use the definition of the de Broglie ‘s wavelength to find the corresponding velocity which in turn will tell us what is a required kinetic energy. Let’s do it

$$
lambda=frac{h}{mv}
$$

So the velocity is given as

$$
v=frac{h}{mlambda}
$$

Now, we can insert this into the expression for kinetic energy

$$
K=frac{mv^2}{2}=frac{mh^2}{2m^2lambda^2}=frac{h^2}{2mlambda^2}
$$

Now, we can plug in the values to have that

$$
K=frac{h^2}{2mlambda^2}=frac{6.63^2times 10^{-68}}{2times 1.67times 10^{-27}times 0.14^2times 10^{-18}}
$$

Which gives us that
$K=671times 10^{-23}textrm{ J}$
And after we transfer it to eVs we have that

$$
K=frac{671times 10^{-23}}{1.6times 10^{-19}}
$$

finally, we have that

$$
boxed{K=0.041textrm{ eV}}
$$

$$
K=0.041textrm{ eV}
$$