JEE Main 2022 — Chemistry Physical Chemistry

De Broglie's equation for calculating the wave length$(\lambda )$(λ) associated with the electron is given below

$\lambda =\frac{h}{p}$λ=ph​$=\frac{h}{\mathrm{mV}}$=mVh​

Let Kinetic energy of the particle of mass 'm' is K

$$K=\frac{1}{2}{\mathrm{mV}}^{2} 2\mathrm{Km}={m}^{2}{V}^{2} \sqrt{2\mathrm{Km}}=\mathrm{mV}$$K=21​mV22Km=m2V22Km​=mV

$$\lambda =\frac{h}{\sqrt{2\mathrm{mK}}}$$λ=2mK​h​

$h$h=Planck's constant=$6.62\times {10}^{-27}\mathrm{erg}-\mathrm{sec}$6.62×10−27erg−sec

$$K=\frac{{h}^{2}}{2{m\lambda }^{2}}$$K=2mλ2h2​

$$K=\frac{{h}^{2}}{2{m\lambda }^{2}}=\frac{43.9\times {10}^{-68}}{2\times 9.1\times {10}^{-31}\times 10.89\times {10}^{-20}}$$K=2mλ2h2​=2×9.1×10−31×10.89×10−2043.9×10−68​

$$K=2.215\times {10}^{-18}$$K=2.215×10−18

${E}_{\mathrm{abs}}={E}_{\mathrm{req}}+K {E}_{\mathrm{abs}}=\mathrm{Energy}\mathrm{absorbed} {E}_{\mathrm{req}=}\mathrm{Energy}\mathrm{required} K=\mathrm{Kinetic}\mathrm{energy}$Eabs​=Ereq​+KEabs​=EnergyabsorbedEreq=​EnergyrequiredK=Kineticenergy$\frac{{E}_{\mathrm{abs}}}{{E}_{\mathrm{req}}}=1+\frac{K}{{E}_{\mathrm{req}}}=1+\frac{2.215\times {10}^{-18}}{13.6\times 1.602\times {10}^{-19}}=2.0166$Ereq​Eabs​​=1+Ereq​K​=1+13.6×1.602×10−192.215×10−18​=2.0166