Answer :
Answer:
0.1 nm
Explanation
Potential deference of the electron is given as V =150 V
Mass of electron [tex]m=9.1\times 10^{-31}[/tex]
Let the velocity of electron = v
Charge on the electron [tex]=1.6\times 10^{-19}C[/tex]
plank's constant h =[tex]6.67\times 10^{-34}[/tex]
According to energy conservation [tex]eV =\frac{1}{2}mv^2[/tex]
[tex]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[/tex]
Now we know that De Broglie wavelength [tex]\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[/tex]
Answer : The De Broglie wavelength of an electron is [tex]1.0025\times 10^{-15}m[/tex]
Explanation :
According to de-Broglie, the expression for wavelength is,
[tex]\lambda=\frac{h}{mv}[/tex]
The formula used for kinetic energy is,
[tex]K.E=\frac{1}{2}mv^2[/tex]
The kinetic energy in terms of momentum will be,
p = m v
[tex]K.E=\frac{1}{2}mv^2=\frac{p^2}{2m}[/tex]
or,
[tex]p=\sqrt{2mK.E}=mv[/tex]
As we know that,
[tex]K.E=q\times V[/tex]
where,
'q' is charge of electron [tex](1.6\times 10^{-9}C)[/tex] and 'V' is potential difference.
So, [tex]p=\sqrt{2m\times q\times V}=mv[/tex]
The de Broglie wavelength of the electron will be:
[tex]\lambda=\frac{h}{\sqrt{2m\times q\times V}}[/tex] .........(1)
where,
h = Planck's constant = [tex]6.626\times 10^{-34}Js=6.626\times 10^{-34}kgm^2/s[/tex]
m = mass of electron = [tex]9.1\times 10^{-31}kg[/tex]
q = charge of electron = [tex]1.6\times 10^{-9}C[/tex]
V = potential difference = 150 V
Now put all the given values in equation 1, we get:
[tex]\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)}}[/tex]
conversion used : [tex](1CV=1J=1kgm^2/s^2)[/tex]
[tex]\lambda=1.0025\times 10^{-15}m[/tex]
Therefore, the De Broglie wavelength of an electron is [tex]1.0025\times 10^{-15}m[/tex]