Units & Dimensions

JEE Main level, single correct, medium

12 of 149

Question 12 of 149

Match List-(I)\left(I\right) with List-(II)\left(II\right). List-(I)\left(I\right) List-(II)\left(II\right) a RH{R}_{H} (Rydberg constant) i kgm−1s−1kg{m}^{-1}{s}^{-1} b hh (Planck's constant) ii kgm2s−1kg{m}^{2}{s}^{-1} c μB{\mu }_{B} (Magnetic field energy density) iii m−1{m}^{-1} d η\eta (coefficient of viscosity) iv kgm−1s−2kg{\text{m}}^{-1}{s}^{-2} Choose the most appropriate answer from the options given below:

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Answer: (C) (a)−(iii),(b)−(ii),(c)−(iv),(d)−(i)(a)-(iii),(b)-(ii),(c)-(iv),(d)-(i)

The relation between wavelength and the principal quantum number is, 1λ=R(1(n1)2−1(n2)2)\dfrac{1}{\lambda }=R\left(\dfrac{1}{{\left({n}_{1}\right)}^{2}}-\dfrac{1}{{\left({n}_{2}\right)}^{2}}\right), for hydrogen-like atom, so the unit to Rydberg constant is, m−1{m}^{-1}.
Now recall the formula of energy of a photon in terms of frequency, E=hνE=h\nu, it means, h=Eν=kgm2s−2s−1=kgm2s−1h=\dfrac{E}{\nu }=\dfrac{kg{m}^{2}{s}^{-2}}{{s}^{-1}}=kg{m}^{2}{s}^{-1}.
The coefficient of viscosity,
η=FA×dvdz=kgms−2mm2×ms−1=kgm−1s−1.\eta =\dfrac{F}{A\times \dfrac{dv}{dz}}=\dfrac{kgm{s}^{-2}m}{{m}^{2}\times m{s}^{-1}}=kg{m}^{-1}{s}^{-1}.
Energy density is defined as, EV=kgm2s−2m3=kgm−1s−2\dfrac{E}{V}=\dfrac{kg{m}^{2}{s}^{-2}}{{m}^{3}}=kg{m}^{-1}{s}^{-2}.

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