The figure shows the variation of photocurrent with anode potential for four different radiations. Let…

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]

1. $f_a=f_b \gt f_c \gt f_d$ and $I_a=I_b \gt I_c \gt I_d$
2. $\mathrm{f}_{\mathrm{a}} \lt \mathrm{f}_{\mathrm{b}} \gt \mathrm{f}_{\mathrm{c}}=\mathrm{f}_{\mathrm{d}}$ and $\mathrm{I}_{\mathrm{a}}=\mathrm{I}_{\mathrm{b}}\gt\mathrm{I}_{\mathrm{c}}^{\prime} \gt \mathrm{I}_{\mathrm{d}}$
3. $\mathrm{f}_{\mathrm{a}}=\mathrm{f}_{\mathrm{b}}=\mathrm{f}_{\mathrm{c}}=\mathrm{f}_{\mathrm{d}}$ and $\mathrm{I}_{\mathrm{a}} \lt \mathrm{I}_{\mathrm{b}} \lt \mathrm{I}_{\mathrm{c}} \lt \mathrm{I}_{\mathrm{d}}$
4. $f_a \gt f_b \gt f_c \gt f_d$ and $I_a=I_b=I_c=I_d$

Solution

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 c, b and a. $\therefore \quad \mathrm{I}_{\mathrm{a}} \lt \mathrm{I}_{\mathrm{b}} \lt \mathrm{I}_{\mathrm{c}} \lt \mathrm{I}_{\mathrm{d}}$ ~

Asked in: MHT CET 2024 (16 May Shift 1)

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