If the energy of a photon corresponding to a wavelength of 6000 $\overset{o}{A}$ is $3 .32 \times 10^{- 19} J$ the photon energy for a wavelength of 4000 $\overset{o}{A}$ will be

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1.4 eV

3.1 eV

1.6 eV

4.9 eV

answer is C.

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Detailed Solution

$\begin{array}{l} E = \frac{hc}{λ} \Rightarrow \frac{E_{1}}{E_{2}} = \frac{λ_{2}}{λ_{1}} \Rightarrow \frac{3 .32 \times 10^{- 19}}{E_{2}} = \frac{4000}{6000} \\ E_{2} = 4 .98 \times 10^{- 19} J = 3 .1 eV \end{array}$

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