First slide

Quantum Theory of Light

Question

If the energy of a photon corresponding to a wavelength of 6000 $Å$ is 3.32 x10^-19joule, the
photon energy (in joule) for a wavelength of 4000 $Å$ will be

Easy

A

1.11 x10^-19
B

2.22 x10-¹⁹
C

4.44 x 10-¹⁹
D

4.98 x 10 -19

Solution

we know that $E = hc / λ$
$E_{1} = \frac{hc}{λ_{1}} and E_{2} = \frac{hc}{λ_{2}} or E_{2} = E_{1} \times (\frac{λ_{1}}{λ_{2}}) = (3 \cdot 32 \times 10^{- 19}) (\frac{6000}{4000}) = (3 \cdot 32 \times 10^{- 19}) (3 / 2) = 4 .98 \times 10^{- 19} joule$

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