Solution:
Since \(\lambda \propto \frac{1}{\sqrt{V}}\), increasing potential by \(300%\) means \(V' = 4V\). Thus, \(\lambda' = \frac{\lambda}{\sqrt{4}} = \frac{\lambda}{2}\).
In an electron microscope electron is accelerated by \(150\text{ kV}\) then de-Broglie wavelength is \(\lambda\). If voltage is increased by \(300\%\) then the de Broglie wavelength of electron will be
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