Question

Calculate the de Broglie wavelength of an electron accelerated from rest through a potential difference of (a) 100 V, (b) 1.0 kV, (c) 100 kV.

Answer 1

Answer:

(a) $\lambda=1.227\ A$

(b) $\lambda=0.388\ A$

(c) $\lambda=0.038\ A$

Explanation:

Given that,

(a) An electron accelerated from rest through a potential difference of 100 V. The De Broglie wavelength in terms of potential difference is given by :

$\lambda=\dfrac{h}{\sqrt{2meV} }$

Where

m and e are the mass of and charge on an electron

On solving,

$\lambda=\dfrac{12.27}{\sqrt{V} }\ A$

V = 100 V

$\lambda=\dfrac{12.27}{\sqrt{100} }\ A$

$\lambda=1.227\ A$

(b) V = 1 kV = 1000 V

$\lambda=\dfrac{12.27}{\sqrt{V} }\ A$

$\lambda=\dfrac{12.27}{\sqrt{1000} }\ A$

$\lambda=0.388\ A$

(c) If $V=100\ kV=10^5\ V$

$\lambda=\dfrac{12.27}{\sqrt{10^5} }\ A$

$\lambda=0.038\ A$

Hence, this is the required solution.