Let initial wavelength be λ, then after 50 increase wavelength will become 1.5λ.
Now, KE=\frac{{P}^{2}}{2m}=eV&P=\frac{h}{\lambda }
Therefore, eV=2m(λh)2.
So we can write
eV1=2m(λh)2...Eq(1) and
eV2=2m(1.5λh)2...Eq(2)
From both equations, we get
V2V1=(1.5)2=49