Test Index

NCERT Class XII Chapter
Dual Nature of Radiation and Matter
Questions With Solutions

© examsnet.com
Question : 13 of 37
Marks: +1, -0
What is the
(a) momentum,
(b) speed, and
(c) de Broglie wavelength of an electron with kinetic energy of 120 eV.
Solution:  
Kinetic energy of electron = 120 × 1.6 × 10−1910^{-19} J = 1.92 × 10−1710^{-17} J
(a) Momentum of electron
p = 2Km\sqrt{2Km} = 2×1.92×10−17×9.1×10−31\sqrt{2 \times 1.92 \times 10^{-17} \times 9.1 \times 10^{-31}} = 5.91 × 10−24 kg m s−110^{-24}\,\mathrm{kg\,m\,s^{-1}}
(b) Speed of electron
v = 2Km\sqrt{\frac{2K}{m}} = 2×1.92×10−179.1×10−31\sqrt{\frac{2 \times 1.92 \times 10^{-17}}{9.1 \times 10^{-31}}} = 6.5 × 10610^{6} m s−1s^{-1}
(c) de-Broglie wavelength associated with electron
λ = hp\frac{h}{p} = 6.63×10−345.91×10−24\frac{6.63 \times 10^{-34}}{5.91 \times 10^{-24}} = 1.122 A0\overset{0}{A}
© examsnet.com
Go to Question:
Prev Question
Next Question