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)

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$