Question

An electron is accelerated through a potential difference of 100V,
then de-Broglie wavelength associated with it is approx.imately​

Answer 1

We have voltage give as 100 V

Work done or Energy = e(x)

and

also Energy= hv

So we have,

e(x) = hv
e(x) = hc/λ

Solving for λ we get,

$λ = \frac{hc}{ex} \\ = \frac{6.626 \times {10}^{ - 34} \times 3x {10}^{8} }{1.6 \times {10}^{ - 19} \times 100 } m \\ \\ = 12.27 \times {10}^{ - 9}m \\ = 1.227angstrom$
Hence your answer.


NOTE: here, e = charge of electron, x = potential difference i.e Voltage, v = frequency of wave, λ is the wavelength, h is plank's constant and c is speed of light.

Answer 2

${\huge{\underline{\underline{\mathcal{\red{♡ANSWER♡}}}}}}$

Acceleration potential ${V=100 v}$ .The de Broglie wavelength $\lambda{is}$

$\lambda{h/p}$=$\frac{1.227}{\sqrt{V}}$ nm

$\lambda$=$\frac{1.227}{\sqrt{100}}$ nm=${0.124nm}$.

The de Broglie wavelength associated with an electron in this case is of the order of X-ray wavelengths.