Question 294: Dual Nature of Matter and Radiation

Question

The figure shows the variation of photocurrent with anode potential for four different radiations. Let $\mathrm{I}_{\mathrm{a}}, \mathrm{I}_{\mathrm{b}}, \mathrm{I}_{\mathrm{c}}$ and $\mathrm{I}_{\mathrm{d}}$ be the intensities for the curves $a, b, c$ and $d$ respectively $\left[f_a, f_b, f_c\right.$ and $f_d$ are frequencies respectively]

Accepted Answer

Since the stopping potential is same, they all have same frequency i.e., $f_a=f_b=f_c=f_d$
From figure Photocurrent is highest for $d$ followed by $\mathrm{c}, \mathrm{b}$ and a.
$$	herefore \quad \mathrm{I}_{\mathrm{a}}<\mathrm{I}_{\mathrm{b}}<\mathrm{I}_{\mathrm{c}}<\mathrm{I}_{\mathrm{d}}$$

Answer

$\mathrm{f}_{\mathrm{a}}=\mathrm{f}_{\mathrm{b}}>\mathrm{f}_{\mathrm{c}}>\mathrm{f}_{\mathrm{d}}$ and $\mathrm{I}_{\mathrm{a}}=\mathrm{I}_{\mathrm{b}}>\mathrm{I}_{\mathrm{c}}>\mathrm{I}_{\mathrm{d}}$

Answer

$\mathrm{f}_{\mathrm{a}}<\mathrm{f}_{\mathrm{b}}>\mathrm{f}_{\mathrm{c}}=\mathrm{f}_{\mathrm{d}}$ and $\mathrm{I}_{\mathrm{a}}=\mathrm{I}_{\mathrm{b}}>\mathrm{I}_{\mathrm{c}}>\mathrm{I}_{\mathrm{d}}$

Answer

$\mathrm{f}_{\mathrm{a}}=\mathrm{f}_{\mathrm{b}}=\mathrm{f}_{\mathrm{c}}=\mathrm{f}_{\mathrm{d}}$ and $\mathrm{I}_{\mathrm{a}}<\mathrm{I}_{\mathrm{b}}<\mathrm{I}_{\mathrm{c}}<\mathrm{I}_{\mathrm{d}}$

Answer

$\mathrm{f}_{\mathrm{a}}>\mathrm{f}_{\mathrm{b}}>\mathrm{f}_{\mathrm{c}}>\mathrm{f}_{\mathrm{d}}$ and $\mathrm{I}_{\mathrm{a}}=\mathrm{I}_{\mathrm{b}}=\mathrm{I}_{\mathrm{c}}=\mathrm{I}_{\mathrm{d}}$

Dual Nature of Matter and Radiation

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