In an experiment on photoelectric effect, a student plots stopping potential V0 against reciprocal of the wavelength λ of the incident light for two different metals A and B. These are shown in the figure. 
Looking at the graphs, you can most appropriately say that:
Held on 25 Apr 2013 · Verified 6 Jul 2026.
Work function of metal B is greater than that of metal A
For light of certain wavelength falling on both metal, maximum kinetic energy of electrons emitted from A will be greater than those emitted from B.
Work function of metal A is greater than that of metal B
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Two p-n junction diodes $D_{1}$ and $D_{2}$ are connected as shown in figure. $A$ and $B$ are input signals and $C$ is the output. The given circuit will function as a $\_\_\_\_$. 
The energy released when $\dfrac{7}{17.13}$ kg of $^{7}_{3}\text{Li}$ is converted into $^{4}_{2}\text{He}$ by proton bombardment is $\alpha \times 10^{32}$ eV. The value of $\alpha$ is _______. (Nearest integer) (Mass of $^{7}_{3}\text{Li} = 7.0183$ u, mass of $^{4}_{2}\text{He} = 4.004$ u, mass of proton $= 1.008$ u and $1$ u $= 931$ MeV/c$^2$ and Avogadro number $= 6.0 \times 10^{23}$)
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. $\left(\mathrm{h}=6.63 \times 10^{-34} \mathrm{~J}. \mathrm{s}, \mathrm{e}=1.6 \times 10^{-19} \mathrm{C}, \mathrm{c}=3 \times 10^{8} \mathrm{~m} / \mathrm{s}\right)$
The de Broglie wavelength of an oxygen molecule at $27^{\circ} \mathrm{C}$ is $x \times 10^{-12} \mathrm{~m}$. The value of $x$ is (take Planck's constant $=6.63 \times 10^{-34} \mathrm{~J}. \mathrm{s}$, Boltzmann constant $=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K}$, mass of oxygen molecule $=5.31 \times 10^{-26} \mathrm{~kg}$)
The given circuit works as : 
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