MCQ
A gas absorbs a photon of $355\, nm$ and emits at two wavelengths. If one of the emissions is at $680\, nm,$ the other is at : ................. $\mathrm{nm}$
- A$1035$
- B$325$
- ✓$743$
- D$518$
According to the law of conservation of energy,
Energy of photon before dividing into two parts $=$ Energy of first photon $+$ Energy of second photon
$\frac{h c}{\lambda}=\frac{h c}{\lambda_{1}}+\frac{h c}{\lambda_{2}}$
$\Rightarrow \frac{1}{\lambda}=\frac{1}{\lambda_{1}}+\frac{1}{\lambda_{2}}$
$\Rightarrow \frac{1}{355}=\frac{1}{680}+\frac{1}{\lambda_{2}}$
$\lambda_{2}=742.77 \mathrm{\,nm}=743 \mathrm{\,nm}$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
At $ 298 K : \Delta_{ f } H ^{\circ}\left( SnO _2( s )\right)=-581.0 kJ mol ^{-1}, \Delta_{ fH } H ^{\circ}\left( CO _2( g )\right)=-394.0 kJ mol ^{-1}$
$S ^{\circ}\left( SnO _2( s )\right)=56.0 J K ^{-1} mol ^{-1}, S ^{\circ}( Sn ( s ))=52.0 J K ^{-1} mol ^{-1}$
$S ^{\circ}( C ( s ))=6.0 J K ^{-1} mol ^{-1}, S ^{\circ}\left( CO _2( g )\right)=210.0 J K ^{-1} mol ^{-1}$
Assume that the enthalpies and the entropies are temperature independent.