Respuesta :
Answer:
Potential difference though which the electron was accelerated is [tex]2.67\times 10^{-6}\ uV\ .[/tex]
Explanation:
Given :
De Broglie wavelength , [tex]\lambda=750\ nm.[/tex]
Plank's constant , [tex]h=6.626\times 10^{-34}\ J.s \ .[/tex]
Charge of electron , [tex]e=-1.6\times 10^{-19}\ C.[/tex]
Mass of electron , m=9.11\times 10^{-31}\ kg.[tex]m=9.11\times 10^{-31}\ kg.[/tex]
We know , according to de broglie equation :
[tex]\lambda=\dfrac{h}{mv}\\\\ v=\dfrac{h}{m\lambda}\\\\v=\dfrac{6.626\times 10^{-34}\ J.s \ }{9.11\times 10^{-31}\ kg\times 750\times 10^{-9}\ m }= 969.78\ m/s .[/tex]
Now , we know potential energy applied on electron will be equal to its kinetic energy .
Therefore ,
[tex]qV=\dfrac{mv^2}{2}\\\\ V=\dfrac{mv^2}{2q}[/tex]
Putting all values in above equation we get ,
[tex]V=2.67\times 10^{-6}\ uV .[/tex]
Hence , this is the required solution.
