Question

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

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

Answer 1

The correct option is B $2\mathring{\text{A}}$

De-Broglie wavelength of a particle is given as,
$\lambda =\frac{h}{p}=\frac{h}{\sqrt{2m(K.E)}}$

Since mass of the particle remains constant
$\lambda \propto \frac{1}{\sqrt{(K.E)}}$

$\Rightarrow \frac{{\lambda }^{\prime }}{\lambda }=\sqrt{\frac{(K.E)}{(K.E{)}^{\prime }}}=\sqrt{\frac{1}{1\times {10}^6}}=\frac{1}{{10}^3}$

$\therefore {\lambda }^{\prime }=\frac{\lambda }{{10}^3}=\frac{2000}{{10}^3}=2\mathring{\text{A}}$

Hence, $\left(B\right)$ is the correct answer.