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(A) The energy of a photon is given by the formula $E = \frac{hc}{\lambda}$.Since $h$ and $c$ are constants,the energy is inversely proportional to th

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$2:3$

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(A) The energy of a photon is given by the formula $E = \frac{hc}{\lambda}$.Since $h$ and $c$ are constants,the energy is inversely proportional to the wavelength: $E \propto \frac{1}{\lambda}$.Therefore,the ratio of energies $E_1 : E_2$ for wavelengths $\lambda_1 = 6000 \ \mathring{A}$ and $\lambda_2 = 4000 \ \mathring{A}$ is:$\frac{E_1}{E_2} = \frac{\lambda_2}{\lambda_1} = \frac{4000}{6000} = \frac{4}{6} = \frac{2}{3}$.Thus,the ratio is $2:3$.

Find the ratio of energy for radiations having $6000 \ \mathring{A}$ and $4000 \ \mathring{A}$ wavelength.
$(h = 6.62 \times 10^{-34} \ J \ s, c = 3 \times 10^8 \ m \ s^{-1})$

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Find the ratio of energy for radiations having $6000 \ \mathring{A}$ and $4000 \ \mathring{A}$ wavelength.$(h = 6.62 \times 10^{-34} \ J \ s, c = 3 \times 10^8 \ m \ s^{-1})$