Explanation:
It is given that,
Voltage, [tex]V=4 eV=4\times 1.6\times 10^{-19}\ V= 6.4\times 10^{-19}\ V[/tex]
De broglie wavelength in terms of voltage is given by :
[tex]\lambda=\dfrac{h}{\sqrt{2meV} }[/tex]
m and e are the mass and charge on electron. So,
[tex]\lambda=\dfrac{12.27}{\sqrt{V} }\ A[/tex]
[tex]\lambda=\dfrac{12.27}{\sqrt{6.4\times 10^{-19}} }\ A[/tex]
[tex]\lambda=1.53\times 10^{10}\ A[/tex]
[tex]\lambda=1.53\ m[/tex]
So, the De broglie wavelength of an electron is 1.53 meters. Hence, this is the required solution.
