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

Question

The de-Broglie wavelength and kinetic energy of a particle is 2000 mathringA and 1 eV respectively. If its kinetic energy becomes 1 MeV, then its de Broglie wavelength (A) 2 mathringA (B) 1 mathringA (C) 4 mathringA (D) 10 mathringA

Tardigrade Admin · Accepted Answer

Given, λ=2000 mathringA KE 1=1 eV and KE 2=1 MeV We know that, λ propto (1/√ KE ) λ × √ KE = text constant Hence, 2000 × √1 eV =λ × √106 eV λ =(2000 × √1 eV /√106 eV ) λ =2 mathringA

Q. The de-Broglie wavelength and kinetic energy of a particle is $2000 \overset{˚}{A}$ and $1 e V$ respectively. If its kinetic energy becomes $1 M e V$ , then its de Broglie wavelength

$2 \overset{˚}{A}$

$1 \overset{˚}{A}$

$4 \overset{˚}{A}$

$10 \overset{˚}{A}$

$5 \overset{˚}{A}$

Given, $λ = 2000 \overset{˚}{A} K E_{1} = 1 e V$ and $K E_{2} = 1 M e V$
We know that,
$λ \propto \frac{1}{K E}$
$λ \times K E = constant$
Hence,
$2000 \times 1 e V = λ \times 1 0^{6} e V$
$λ = \frac{2000 \times 1 e V}{1 0 ^{6} e V}$
$λ = 2 \overset{˚}{A}$

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

2A˚

1A˚

4A˚

10A˚

5A˚

Given, λ=2000A˚KE1​=1eV and KE2​=1MeV We know that, λ∝KE​1​ λ×KE​= constant Hence, 2000×1eV​=λ×106eV​ λ=106eV​2000×1eV​​ λ=2A˚