The de Broglie wavelength of an electron accelerated through a potential difference V is given by λ=2meVh.
This implies λ∝V1.
Given that the new wavelength λ2 is increased by 50%, we have λ2=λ1+0.5λ1=23λ1.
Taking the ratio of the wavelengths, we get λ2λ1=V1V2.
Substituting λ2=23λ1, we get 32=V1V2.
Squaring both sides yields V1V2=94, which gives V2V1=49.
Comparing this with the given relation V2V1=α9, we find α=4.
Answer: 4