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The wavelengths involved in the spectrum of deuterium (12D) are slightly different from that of hydrogen spectrum, because
Held on 30 Apr 2003 · Verified 6 Jul 2026.
the size of the two nuclei are different
the nuclear forces are different in the two cases
the masses of the two nuclei are different
the attraction between the electron and the nucleus is different in the two cases
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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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