Question 2The frequency range of visible light is from 3.75 × 1014 Hz to 7.5 × 1014 Hz. Calculate its wavelength range. Take speed of light = 3 x 108 m s-1

Question

Accepted Answer

As we know,Speed of light (c) = Frequency (f) x Wavelength (λ)Given,C = 3 x 108 m s-1Frequency range = 3.75 × 1014 Hz to 7.5 × 1014 Hz.Let, f1 = 3.75 × 1014 HzSubstituting the values in the formula,3×108=3.75×1014×λλ=3×1083.75×1014⇒λ=0.8×10−6m⇒λ=8000×10−10m3 \times 10^8 = 3.75 × 10^{14} \times λ \[0.5em] \text {λ} = \dfrac {3 \times 10^8 }{3.75 × 10^{14}} \[0.5em] \Rightarrow \text {λ} = 0.8 \times 10^{-6} \text {m} \[0.5em] \Rightarrow \text {λ} = 8000 \times 10^{-10} \text {m} \[0.5em]3×108=3.75×1014×λλ=3.75×10143×108​⇒λ=0.8×10−6m⇒λ=8000×10−10mAs, 1 Å = 10-10 mTherefore,λ1 = 8000 ÅLet, f2 = 7.5 × 1014 HzSubstituting the values in the formula,3×108=7.5×1014×λλ=3×1087.5×1014⇒λ=0.4×10−6m⇒λ=4000×10−10m3 \times 10^8 = 7.5 × 10^{14} \times λ \[0.5em] \text {λ} = \dfrac {3 \times 10^8 }{7.5 \times 10^{14}} \[0.5em] \Rightarrow \text {λ} = 0.4 \times 10^{-6} \text {m} \[0.5em] \Rightarrow \text {λ} = 4000 \times 10^{-10} \text {m} \[0.5em]3×108=7.5×1014×λλ=7.5×10143×108​⇒λ=0.4×10−6m⇒λ=4000×10−10mAs, 1 Å = 10-10 mTherefore,λ2 = 4000 ÅTherefore, the wavelength range is 4000 Å to 8000 Å.

Spectrum — Question 2

Question 2