Question

A nonrelativistic electron is accelerated from rest through a potential difference. After acceleration the electron has a de Broglie wavelength of 750 nm. What is the potential difference though which this electron was accelerated? (h = 6.626 × 10-34 J · s, e = - 1.60 × 10-19 C, mel = 9.11 × 10-31 kg)

Answer 1

Answer:

Potential difference though which the electron was accelerated is $2.67\times 10^{-6}\ uV\ .$

Explanation:

Given :

De Broglie wavelength , $\lambda=750\ nm.$

Plank's constant , $h=6.626\times 10^{-34}\ J.s \ .$

Charge of electron , $e=-1.6\times 10^{-19}\ C.$

Mass of electron , m=9.11\times 10^{-31}\ kg.$m=9.11\times 10^{-31}\ kg.$

We know , according to de broglie equation :

$\lambda=\dfrac{h}{mv}\\\\ v=\dfrac{h}{m\lambda}\\\\v=\dfrac{6.626\times 10^{-34}\ J.s \ }{9.11\times 10^{-31}\ kg\times 750\times 10^{-9}\ m }= 969.78\ m/s .$

Now , we know potential energy applied on electron will be equal to its kinetic energy .

Therefore ,

$qV=\dfrac{mv^2}{2}\\\\ V=\dfrac{mv^2}{2q}$

Putting all values in above equation we get ,

$V=2.67\times 10^{-6}\ uV .$

Hence , this is the required solution.