The longest wavelength doublet absorption transition is observed at 589 and 589.6 nm. Energy difference between two excited states is

Question

The longest wavelength doublet absorption transition is observed at 589 and 589.6 nm. Energy difference between two excited states is (A) 3.31 × 10-22 kJ (B) 3.31 × 10-22 J (C) 2.98 × 10-21 J (D) 3.0 × 10-21 kJ

Tardigrade Admin · Accepted Answer

Step I: Wavelength (λ) = 589 nm = 589 × 10-9 m Frequency ( upsilon)=(c/λ)=(3 × 108/589 × 10-9) =5.093×1014 cycles per sec Step II: Wavelength (λ)=589.6 nm=589.6×10-9 m ∴ upsilon=(c/λ)=(3 × 108/589.6 ×) 10-9 =5.088×1014 cycles per sec Energy difference between two excited states, Δ E=6.626×10-34 (5.093-5.088)1014 Δ x.mΔ v=(h/4π) Uncertainty in the position, Δ x=(h/4π mΔ v) =(6.626 × 10-34/4 × 3.14 × 10 × 10-3 × 10) =5.27×10-34 m

Q. The longest wavelength doublet absorption transition is observed at $589$ and $589.6 nm$ . Energy difference between two excited states is

$3.31 \times 1 0^{- 22} k J$

$3.31 \times 1 0^{- 22} J$

$2.98 \times 1 0^{- 21} J$

$3.0 \times 1 0^{- 21} k J$

Q. The longest wavelength doublet absorption transition is observed at 589 and 589.6nm. Energy difference between two excited states is

3.31×10−22kJ

3.31×10−22J

2.98×10−21J

3.0×10−21kJ

Q. The longest wavelength doublet absorption transition is observed at $589$ and $589.6 nm$ . Energy difference between two excited states is

$3.31 \times 1 0^{- 22} k J$

$3.31 \times 1 0^{- 22} J$

$2.98 \times 1 0^{- 21} J$

$3.0 \times 1 0^{- 21} k J$