Explanation:
The relationship between the wavelength and the potential difference V is given by :
[tex]\lambda=\dfrac{h}{\sqrt{2me}}\times \dfrac{1}{\sqrt{V}}[/tex]
Putting the values of known parameters,
[tex]\lambda=\dfrac{12.28}{\sqrt{V} }\times 10^{-10}\ m[/tex]
(a) [tex]V=30\ kV=30000\ V[/tex]
[tex]\lambda=\dfrac{12.28}{\sqrt{30000} }\times 10^{-10}\ m[/tex]
[tex]\lambda=7.08\times 10^{-12}\ m[/tex]
(b) [tex]V=60\ kV=60000\ V[/tex]
[tex]\lambda=\dfrac{12.28}{\sqrt{60000} }\times 10^{-10}\ m[/tex]
[tex]\lambda=5.01\times 10^{-12}\ m[/tex]
Hence, this is the required solution.
