# A nonrelativistic electron is accelerated from rest through a potential difference. After acceleration the electron has a de Broglie wavelen

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)

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1 year 2021-08-31T14:41:32+00:00 1 Answers 16 views 0

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.