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NCERT Class XII Chapter
Dual Nature of Radiation and Matter
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Question : 12 of 37
Marks: +1, -0
Calculate the
(a) momentum, and
(b) de Broglie wavelength of the electrons accelerated through a potential difference of 56 V.
Solution:  
An electron which is accelerated through a potential difference of 56 V will have kinetic energy K = 56 eV
(a) Momentum associated with accelerated electron
P = 2Km\sqrt{2Km} =
2×56×1.6×10−19×9.1×10−31\sqrt{2 \times 56 \times 1.6 \times 10^{-19} \times 9.1 \times 10^{-31}}
= 4.04 × 10−2410^{-24} Kg m s−1s^{-1}
(b) Wavelength of electron accelerated
λ = hp\frac{h}{p} = 6.63×10−344.04×10−24\frac{6.63 \times 10^{-34}}{4.04 \times 10^{-24}} = 0.164 nm
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