Question

Calculate the de-Broglie wavelength of an electron and an alpha particle accelerated with the same potential of 100 V.
Solution:

Answer 1

The de-Broglie wavelength of the electron is  0.123 nm

Explanation:

The de-Broglie wavelength "w" for a particle with momentum "p" is given by the formula.

w = h/p,

where h is Plank’s constant.

To calculate energy of electron we can use formula.

• W = V x q

Here "w" is energy transfer.

q = charge

W = E = K.E

W = 100 eV

W = 1.6 x 10^-17 J (100 x e)

• Now using another equation.

E = p^2/2 x m

where m is the mass of the electron.

p = (2.m.E)^1/2

p = (2 x 9.1 x 10^-31 x 1.6 x 10^-17)^1/2

p = 5.39 x 10^-24 kgm/s

Now

w = h/p = 6.63 x 10^-34/5.39 x 10^-24

= 1.23 x 10^-10 m

= 0.123 nm

Thus the de-Broglie wavelength of the electron is  0.123 nm

Also learn more

Calculate the wavelength of a particle of mass m=6.62*10^-27 kg moving with kinetic energy 7.425*10^-13 joule ?

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