The de Broglie wavelength and kinetic energy of a particle is 2000 Å and 1 eV respectively. If its kinetic energy becomes 1 MeV, then its de Broglie wavelength becomes

2 Å

As $λ=\frac{h}{\sqrt{2mK}}$

Since mass of the particle remains constant

$∴λ∝\frac{1}{\sqrt{K}}$

$\frac{λ'}{λ}=\sqrt{\frac{K}{K'}}=\sqrt{\frac{1eV}{1×10^6eV}}=\frac{1}{10^3}$

or $λ'=\frac{λ}{10^3}=\frac{2000}{10^3}Å=2Å$