Answer:
De broglie wavelength will be 0.727 nm
Step-by-step explanation:
We have given speed of the electron
![v=10^6m/sec](https://img.qammunity.org/2020/formulas/physics/college/262s7xvqluvtwf74yruvvmw3stdua47lz3.png)
Mass of the electron
![m=9.11* 10^(-31)kg](https://img.qammunity.org/2020/formulas/engineering/college/4fx8xs091cfo1xk86m50zsmm56ateamo12.png)
Plank's constant
![h=6.624* 10^(-34)Js](https://img.qammunity.org/2020/formulas/physics/college/brr8atsfuvhz753iqvca8rb1xekx96eqf2.png)
We have to find the De Broglie wavelength
De Broglie wavelength is given by
![\lambda =(h)/(mv)=(6.624* 10^(-34))/(9.11* 10^(-31)* 10^6)=0.727* 10^(-9)m](https://img.qammunity.org/2020/formulas/physics/college/w6i24zkiclvxndqk2up7e5tt8w61jp4c90.png)
So wavelength will be 0.727 nm