Dual Nature of Radiation and Matter - Result Question 18

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19. The wavelength associated with an electron, accelerated through a potential difference of 100 $V$, is of the order of

======= ####19. The wavelength associated with an electron, accelerated through a potential difference of 100 $V$, is of the order of

3e0f7ab6f6a50373c3f2dbda6ca2533482a77bed:content/english/neet-pyq-chapterwise/physics/dual-nature-of-radiation-and-matter/dual-nature-of-radiation-and-matter—result-question-18.md (a) $1000 \AA$

(b) $100 \AA$

(c) $10.5 \AA$

(d) $1.2 \AA$

[1996]

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Answer:

Correct Answer: 19. (d)

Solution:

  1. (d) Potential difference $=100 V$

K.E. acquired by electron $=e(100)$

$ \frac{1}{2} m v^{2}=e(100) \Rightarrow v=\sqrt{\frac{2 e(100)}{m}} $

According to de Broglie’s concept

$ \begin{aligned} & \lambda=\frac{h}{m \nu} \quad \Rightarrow \lambda=\frac{h}{m \sqrt{\frac{2 e(100)}{m}}} \\ & =\frac{h}{\sqrt{2 m e(100)}}=1.2 \times 10^{-10}=1.2 \AA \end{aligned} $

de-Broglie wavelength, $\lambda_e=\frac{h}{p}=\frac{h}{\sqrt{2 m E}}$

Here, $h=$ planck’s constant

$m=$ mass of electron

Kinetic energy, $E=eV$

Here, $V=$ potential difference

$ \therefore \lambda_e \frac{h}{\sqrt{2 m_e e V}} $

Substituting h $=6.63 \times 10^{-34} Js$

$e=1.6 \times 10^{-19} C$

$m_e=9.1 \times 10^{-31} kg$

we get $\lambda_e=\frac{12.27}{\sqrt{V}} A$