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Arrive at the expression for the wavelen...

Arrive at the expression for the wavelength of electron in terms of electric potential.

Text Solution

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We know that the wavelength associated with the accelerating electron,
i.e. `lambda_(e) = h/sqrt(2emV) = (12.27) /sqrt(V)` Å
Hence `lambda_(e) prop 1/sqrt(V)` where `V =` is the electric potential.
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