A photon detector captures a photon with an energy of 5.45 ✕ 10^−19 J. What is the wavelength, in nanometers, of the photon?

Chemistry

1 answer:

GarryVolchara [31]3 years ago

5 0

Answer:

364 nm

Explanation:

The formula for the energy (E) of a photon in terms of its wavelength λ is

E = hc/λ

We can rearrange this formula to give

λ = hc/E

If E = 5.45 × 10⁻¹⁹ J, then

$\begin{array}{rcl}\lambda& = & \dfrac{6.626 \times 10^{-34} \text{ J$\cdot$s} \times 2.998 \times 10^{8}\text{ m$\cdot$ s}^{-1}} {5.45 \times 10^{19} \text{ J}}\\\\& = & 3.64 \times 10^{-7} \text{ m}\\& = & \textbf{364 nm}\\\end{array}\\$

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