The de Broglie wavelength of an electron in free space is given by λ0=mvh.
When the electron enters the medium, its velocity is reduced by 20%. The new velocity is v′=v−0.20v=0.80v.
The de Broglie wavelength in the medium is λ′=mv′h.
Substituting v′=0.80v, we get λ′=m(0.80v)h=0.801(mvh).
Since \dfrac{1}{0.80} = \dfrac{100}{80} = 1.25,wehave\lambda' = 1.25\lambda_0$.
Comparing this with λ′=αλ0, we get α=1.25.
Answer: 1.25