Energy of a photon with a wavelength of 450 nm is

# Energy of a photon with a wavelength of 450 nm is

1. A

$4.36×{10}^{-12}\mathrm{ergs}$

2. B

$4.36×{10}^{-13}\mathrm{ergs}$

3. C

$4.36×{10}^{-20}\mathrm{ergs}$

4. D

$4.36×{10}^{-11}\mathrm{ergs}$

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### Solution:

According to Max Planck’s equation,

$\mathrm{E}=\frac{\mathrm{hc}}{\mathrm{\lambda }}\to \mathrm{eqn}\left(1\right)$

Here h is Planck’s constant and c is speed of light.

The wave length of photon $\mathrm{\lambda }$ = 450nm

$\mathrm{h}=6.625×{10}^{-27}\mathrm{ergs}.\mathrm{sec}$

$\mathrm{c}=3×{10}^{10}\mathrm{cm}/\mathrm{sec}$

From equation 1, the energy of photon is

$\mathrm{E}=\frac{6.625×{10}^{-27}\mathrm{ergs}.\mathrm{sec}×3×{10}^{10}\mathrm{cm}.{\mathrm{sec}}^{-1}}{450×{10}^{-7}\mathrm{cm}}$$\therefore \mathrm{E}=4.36×{10}^{-12}\mathrm{ergs}$

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