Dual Nature of Radiation and Matter - Result Question 20
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21. An ionization chamber with parallel conducting plates as anode and cathode has $5 \times 10^{7}$ electrons and the same number of singly charged positive ions per $cm^{3}$. The electrons are moving towards the anode with velocity $0.4 m / s$. The current density from anode to cathode is $4 \mu A$ / $m^{2}$. The velocity of positive ions moving towards cathode is
======= ####21. An ionization chamber with parallel conducting plates as anode and cathode has $5 \times 10^{7}$ electrons and the same number of singly charged positive ions per $cm^{3}$. The electrons are moving towards the anode with velocity $0.4 m / s$. The current density from anode to cathode is $4 \mu A$ / $m^{2}$. The velocity of positive ions moving towards cathode is
3e0f7ab6f6a50373c3f2dbda6ca2533482a77bed:content/english/neet-pyq-chapterwise/physics/dual-nature-of-radiation-and-matter/dual-nature-of-radiation-and-matter—result-question-20.md (a) $0.4 m / s$
(b) $1.6 m / s$
(c) zero
(d) $0.1 m / s$
[1992]
Topic 2: Electron Emission, Photon
Photoelectric Effect & X-ray22. Light of frequency 1.5 times the threshold frequency is incident on a photosensitive material. What will be the photoelectric current if the frequency is halved and intensity is doubled?
(a) four times
(b) one-fourth
(c) zero
(d) doubled
[2020]
Show Answer
Answer:
Correct Answer: 21. (d)
Solution:
- (d) Current $=I_e+I_p$
$I_e$ and $I_p$ are current due to electrons and positively charged ions.
$I=n e A V_d$
where,
$ \begin{aligned} n=5 \times 10^{7} / cm^{3}= & 5 \times 10^{7} \times 10^{6} / m^{3} \\ & =5 \times 10^{13} / m^{3} \end{aligned} $
$I_e=5 \times 10^{13} \times 1.6 \times 10^{-19} \times A \times 0.4$
$I_p=5 \times 10^{13} \times 1.6 \times 10^{-19} \times A \times v$
$I=I_e+I_p$ (from equation (i))
$=5 \times 10^{13} \times 1.6 \times 10^{-19} \times A(v+0.4)$
Given, $I / A=4 \times 10^{-6} A / m^{2}$
$4 \times 10^{-6} \times A=5 \times 10^{-6} \times 1.6 \times A(v+0.4)$
$\frac{4}{8}=v+0.4 \Rightarrow 0.5=v+0.4 \Rightarrow v=0.1 m / s$
