The formula of De Broglie is the following:
[tex]\lambda=\frac{h}{mv}\begin{cases}\lambda=wavelength\text{ (m)} \\ h=Planck\text{ constant =6.626}\cdot10^{-34}Js \\ m=mas\text{s (kg)} \\ \text{v=}velocity\text{ (m/s)}\end{cases},[/tex]
Remember that the mass of an electron is 9.11 x 10 ^(-31) kg, so replacing in the formula, we're going to obtain:
[tex]\begin{gathered} \lambda=\frac{6.626\cdot10^{-34}J\cdot s}{9.11\cdot10^{-31}\operatorname{kg}\cdot1.48\cdot10^5\frac{m}{s}}, \\ \lambda=4.91\cdot10^{-9}\text{ m} \end{gathered}[/tex]
The answer is that the wavelength of an electron traveling at 1.48 x 10^(5) m/s is 4.91 x 10^(-9) m.
