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

Options

A. a)2 Å
B. b)1 Å
C. c)4 Å
D. d)10 Å

Answer

A

Explanation

Given, `lambda = 2000 Å, KE_1 = 1e V and KE_2 = 1 MeV`
We know that,
`lambda prop (1)/( sqrt(KE))`
`lambda xx sqrt(KE) =` constant
Hence,
`2000 xx sqrt(1 eV) = lambda xx sqrt(10^(6) eV)`
`lambda = (2000 xx sqrt(1eV))/(sqrt( 10^(6) eV))`
`lambda =2 Å`