KCET · Physics · Dual Nature of Matter
The following graph represents the variation of photocurrent with anode potential for a metal surface. Here \(I_{1}, I_{2}\) and \(I_{3}\) represents intensities and \(\gamma_{1}, \gamma_{2}, \gamma_{3}\) represent frequency for curves 1,2 and 3 respectively, then
- A \(\gamma_{1}=\gamma_{2}\) and \(I_{1} \neq I_{2}\)
- B \(\gamma_{1}=\gamma_{3}\) and \(I_{1} \neq I_{3}\)
- C \(\gamma_{1}=\gamma_{2}\) and \(I_{1}=I_{2}\)
- D \(\gamma_{2}=\gamma_{3}\) and \(I_{1}=I_{3}\)
Answer & Solution
Correct Answer
(A) \(\gamma_{1}=\gamma_{2}\) and \(I_{1} \neq I_{2}\)
Step-by-step Solution
Detailed explanation
The given figure can be shown as
From figure, we can see that the stopping potential \(\left(V_{0}\right)\) is same for curve 1 and 2. But for curve 3 , it is greater than that for land 2 .
As, we know that, \(e V_{0}=E_{\max }=h \gamma-\phi_{0}\)
where, \(\phi_{0}\) is the work function.
Since, \(V_{0}\) is same for 1 and 2, so from above equation, we have \(r_{1}=r_{2}\)
Thus, for 1,2 and 3
\(\gamma_{1}=\gamma_{2} \neq \gamma_{3}\)
Also, the saturation current of curve 2 is greater than of curve 1 . But for curve 2 and 3 it is equal, so,
\(I_{1} \neq I_{2}=I_{3}\)
From figure, we can see that the stopping potential \(\left(V_{0}\right)\) is same for curve 1 and 2. But for curve 3 , it is greater than that for land 2 .
As, we know that, \(e V_{0}=E_{\max }=h \gamma-\phi_{0}\)
where, \(\phi_{0}\) is the work function.
Since, \(V_{0}\) is same for 1 and 2, so from above equation, we have \(r_{1}=r_{2}\)
Thus, for 1,2 and 3
\(\gamma_{1}=\gamma_{2} \neq \gamma_{3}\)
Also, the saturation current of curve 2 is greater than of curve 1 . But for curve 2 and 3 it is equal, so,
\(I_{1} \neq I_{2}=I_{3}\)
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