When a light of a given wavelength falls on a metallic surface the stopping potential for photoelectrons is $3.2 \ V$. If a second light having wavelength twice of first light is used, the stopping potential drops to $0.7 \ V$. The wavelength of first light is . . . . . . $m$.
$(h = 6.63 \times 10^{-34} \ J.s, e = 1.6 \times 10^{-19} \ C, c = 3 \times 10^8 \ m/s)$
$(h = 6.63 \times 10^{-34} \ J.s, e = 1.6 \times 10^{-19} \ C, c = 3 \times 10^8 \ m/s)$
- A$2.9 \times 10^{-8}$
- B$2.2 \times 10^{-8}$
- C$3.1 \times 10^{-7}$
- D$2.5 \times 10^{-7}$
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