Dual Nature of Radiation and Matter - Result Question 18
19. The wavelength associated with an electron, accelerated through a potential difference of 100 $V$, is of the order of
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======= ####19. The wavelength associated with an electron, accelerated through a potential difference of 100 $V$, is of the order of
c3eec34ec6b1fad69db54a20ad4b2dca40c2aa54 (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:
- (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 $\lambda_e=\frac{12.27}{\sqrt{V}} A$
