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CBSE Questions for Class 12 Medical Physics Dual Nature Of Radiation And Matter Quiz 16 - MCQExams.com
CBSE
Class 12 Medical Physics
Dual Nature Of Radiation And Matter
Quiz 16
An electron (of mass m) and a photon have the same energy E in the range of a few eV. The ratio of the de-Broglie wavelength associated with the electron and the wavelength of the photon is (c = speed of light in vacuum)
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$$\dfrac{1}{c} \left(\dfrac{E}{2m} \right)^{1/2}$$
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$$c (2mE)^{1/2}$$
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$$\left(\dfrac{E}{2m} \right)^{1/2}$$
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$$\dfrac{1}{c} \left(\dfrac{2E}{m} \right)^{1/2}$$
Explanation
For electron
De - Broglie wavelength $$\lambda_c = \dfrac{h}{p}$$
where p is momentum $$p = mv$$
also by energy we have $$E = \dfrac{1}{2} mv^2$$
$$\Rightarrow E = \dfrac{1}{2} \dfrac{p^2}{m}$$
$$\Rightarrow p = \sqrt{2 mE}$$
$$\therefore \lambda_c = \dfrac{h}{\sqrt{2 mE}}$$
For photon energy $$\Rightarrow E = \dfrac{hc}{\lambda}$$
$$\Rightarrow \lambda = \dfrac{h c}{E}$$
$$\therefore \dfrac{\lambda_c}{\lambda} = \dfrac{h}{\sqrt{2 mE}} \dfrac{E}{hc}$$
$$= \dfrac{1}{C} \sqrt{\dfrac{E}{2m}}$$
option (A) is correct.
A particle moving with kinetic energy E has de Broglie wavelength $$\lambda$$. If energy $$\Delta E$$ is added to its energy, the wavelength become $$\lambda /2$$. Value of $$\Delta E$$, is?
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$$2E$$
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$$4E$$
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$$3E$$
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$$E$$
Explanation
For Partial,
$$\lambda=\dfrac{h}{\sqrt{2\times (K.E)\times m}}$$
$$\lambda \propto \dfrac{1}{\sqrt{K.E}}$$
$$\dfrac{\lambda_1}{\lambda_2}=\sqrt{\dfrac{(K.E)_2}{(K.E)_1}}=\sqrt{\left(\dfrac{E+\Delta E}{E}\right)}$$
$$\left(\dfrac{\lambda}{\lambda/2}\right)^2=\dfrac{E+\Delta E}{E}$$
$$4=1+\dfrac{\Delta E}{E}$$
$$3=\dfrac{\Delta E}{E}$$
$$\boxed{\Delta E=3E}$$
Option (C) is correct.
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