Question

The de-Broglie wavelength and kinetic energy of a particle is $2000\mathring{A}$ and 1 eV respectively. If its kinetic energy becomes 1 MeV, then its de-Broglie wavelength is

• A. $2\mathring{A}$
• B. $1\mathring{A}$
• C. $4\mathring{A}$
• D. $10\mathring{A}$
• E. $5\mathring{A}$

Answer 1

Answer: (A)

$2\mathring{A}$

The de-Broglie wavelength, $\lambda =\frac{h}{p}$

where $h=$ Plank's constant and $p=$ momentum of the particle

Kinetic energy, $E=\frac{p^2}{2m}=\frac{h^2}{2m{\lambda }^2}$

As mass m and h are constants in both case, so $E\propto \frac{1}{{\lambda }^2}$

Thus, $\frac{E_1}{E_2}=\frac{{\lambda }_2^2}{{\lambda }_1^2}$

or ${\lambda }_2^2=\left(E_1/E_2\right){\lambda }_1^2=\frac{1eV}{1\times {10}^{6}eV}(2000A^o{)}^2$

or ${\lambda }_2=2A^o$