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NCERT Class XII Chapter
Dual Nature of Radiation and Matter
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Question : 14 of 37
Marks: +1, -0
The wavelength of light from the spectral emission line of sodium is 589 nm. Find the kinetic energy at which
(a) an electron, and
(b) a neutron, would have the same de Broglie wavelength.
Solution:  
(a) Kinetic energy required by electron to have de-Broglie wavelength of 589 nm
K = h22mλ2\frac{h^2}{2m\lambda^2} = (6.63×10−34)22×9.1×10−31×(589×10−9)2\frac{(6.63 \times 10^{-34})^2}{2 \times 9.1 \times 10^{-31} \times (589 \times 10^{-9})^2}
K = 6.95 × 10−2510^{-25} J = 4.34 µeV
(b) Kinetic energy of neutron to have de-Broglie wavelength of 589 nm
K = h22mλ2\frac{h^2}{2m\lambda^2} = (6.63×10−34)22×1.67×10−27×(589×10−9)2\frac{(6.63 \times 10^{-34})^2}{2 \times 1.67 \times 10^{-27} \times (589 \times 10^{-9})^2}
K = 3.78 × 10−2810^{-28} J = 2.36 neV
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