Question

The de Broglie wavelength of an electron is $0.4 � 10^{-10}$ m when its kinetic energy is $1.0\, keV$. Its wavelength will be $1.0 \times 10^{-10}\, m$, when its kinetic energy is

• A. 0.2 keV
• B. 0.8 keV
• C. 0.63 keV
• D. 0.16 keV

Answer 1

Answer: (D)

0.16 keV

Answer 2

de-Broglie wavelength,
$\lambda=\frac{ h }{\sqrt{2 m E}}$
$\therefore \frac{\lambda_{1}}{\lambda_{2}} =\sqrt{\frac{E_{2}}{E_{1}}}$
$\lambda_{1} =$ Wave length of electron
$=0 \cdot 4 \times 10^{-10} m$
$E_{1} =1 \cdot 0\, keV$
$\lambda_{2} =1 \cdot 0 \times 10^{-10} m$
$E_{2} =?$
$\frac{0.4 \times 10^{-10}}{1 \times 10^{-10}} =\sqrt{\frac{E_{2}}{1 \cdot 0 keV }}$
$\frac{4}{10} =\sqrt{\frac{E_{2}}{1}}$
$\frac{16}{100} =\frac{E_{2}}{1}$
$E_{2} =0 \cdot 16 \,keV$