Question

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

Answer 1

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 $1.0025\times 10^{-15}m$

Answer 2

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 $\lambda =\frac{h}{mv}=\frac{6.67\times 10^{-34}}{9.1\times 10^{-31}\times 7.2627\times10^6 }=0.100\times 10^{-9}m=0.1nm$