Explanation:
(a) An electron accelerated from rest through a potential difference of 100 V. The De Broglie wavelength in terms of potential difference is given by :
[tex]\lambda=\dfrac{h}{\sqrt{2meV} }[/tex]
Where
m and e are the mass of and charge on an electron
Here, eV = 50 eV
[tex]\lambda=\dfrac{6.67\times 10^{-34}}{\sqrt{2\times 9.1\times 10^{-31}\times 50} }[/tex]
[tex]\lambda=6.99\times 10^{-20}\ m[/tex]
The frequency of electron is given by :
[tex]f=\dfrac{c}{\lambda}[/tex]
[tex]f=\dfrac{3\times 10^8}{6.99\times 10^{-20}}[/tex]
[tex]f=4.29\times 10^{27}\ Hz[/tex]
(b) The mass of the proton, [tex]m=1.67\times 10^{-27}\ kg[/tex]
The De Broglie wavelength is given by :
[tex]\lambda=\dfrac{6.67\times 10^{-34}}{\sqrt{2\times 1.67\times 10^{-27}\times 50}}[/tex]
[tex]\lambda=1.63\times 10^{-21}\ m[/tex]
The frequency of proton is given by :
[tex]f=\dfrac{c}{\lambda}[/tex]
[tex]f=\dfrac{3\times 10^8}{1.63\times 10^{-21}}[/tex]
[tex]f=1.84\times 10^{29}\ Hz[/tex]
Hence, this is the required solution.
