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MAVERICK [17]
3 years ago
13

What is the De Broglie wavelength of an electron under 150 V acceleration?

Engineering
2 answers:
SVEN [57.7K]3 years ago
7 0

Answer : The De Broglie wavelength of an electron is 1.0025\times 10^{-15}m

Explanation :

According to de-Broglie, the expression for wavelength is,

\lambda=\frac{h}{mv}

The formula used for kinetic energy is,

K.E=\frac{1}{2}mv^2

The kinetic energy in terms of momentum will be,

p = m v

K.E=\frac{1}{2}mv^2=\frac{p^2}{2m}

or,

p=\sqrt{2mK.E}=mv

As we know that,

K.E=q\times V

where,

'q' is charge of electron (1.6\times 10^{-9}C) and 'V' is potential difference.

So, p=\sqrt{2m\times q\times V}=mv

The de Broglie wavelength of the electron will be:

\lambda=\frac{h}{\sqrt{2m\times q\times V}}   .........(1)

where,

h = Planck's constant = 6.626\times 10^{-34}Js=6.626\times 10^{-34}kgm^2/s

m = mass of electron = 9.1\times 10^{-31}kg

q = charge of electron = 1.6\times 10^{-9}C

V = potential difference = 150 V

Now put all the given values in equation 1, we get:

\lambda=\frac{6.626\times 10^{-34}kgm^2/s}{\sqrt{2\times (9.1\times 10^{-31}kg)\times (1.6\times 10^{-9}C)\times (150V)}}

conversion used : (1CV=1J=1kgm^2/s^2)

\lambda=1.0025\times 10^{-15}m

Therefore, the De Broglie wavelength of an electron is 1.0025\times 10^{-15}m

yanalaym [24]3 years ago
4 0

Answer:

0.1 nm

Explanation

Potential deference of the electron is given as V =150 V

Mass of electron m=9.1\times 10^{-31}

Let the velocity of electron = v

Charge on the electron =1.6\times 10^{-19}C

plank's constant h =6.67\times 10^{-34}

According to energy conservation eV =\frac{1}{2}mv^2

v=\sqrt{\frac{2eV}{M}}=\sqrt{\frac{2\times 1.6\times 10^{-19}\times 150}{9.1\times 10^{-31}}}=7.2627\times 10^{-6}m/sec

Now we know that De Broglie wavelength \lambda =\frac{h}{mv}=\frac{6.67\times 10^{-34}}{9.1\times 10^{-31}\times 7.2627\times10^6 }=0.100\times 10^{-9}m=0.1nm

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