Question

Calculate the de-Broglie wavelength of an alpha particle accelerated through potential difference of 100V.​

Answer 1

Given:

Potential difference,

V = 100

To find:

Wavelength = ?

Solution:

→ The wavelength of electrons will be:

= $\frac{12.27}{\sqrt{V} }A$

By putting the values, we get

= $\frac{12.27}{\sqrt{100} }$

= $1.227 \ A$

Now,

→ $q_a = +2e$

       $=2\times 1.6\times 10^{-19}$

       $=3.2\times 10^{-19} \ C$

For α-particle,

→ The wavelength (λ) will be:

= $\frac{h}{\sqrt{2mqV} }$

= $\frac{6.63\times 10^{-34}}{\sqrt{2\times 4\times 1.67\times 10^{-27}\times 3.2\times 10^{-19}\times 100} }$

= $0.010 \ A$

Thus the wavelength of an alpha will be "0.010 A".

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