Question

Question #02253

Answer 1

`E = 3.03 xx 10^-20` `"J"`

Explanation:

We're asked to calculate the energy if a photon given its wavelength.

To do this, we can use the equation

`E = (hc)/lambda`

where

• `E` is the energy of the photon, in `"J"`

• `h` is Planck's constant, equal to `6.626 xx 10^-34` `"J"•"s"`

• `c` is the speed of light in vacuum, equal to `299792458` `"m/s"`

• `lambda` is the wavelength of the photon, in `"m"`

We therefore need to convert from `"μm"` to `"m"`:

`6.55cancel("μm")((1color(white)(l)"m")/(10^6cancel("μm"))) = 6.55 xx 10^-6` `"m"`

Plugging in known values, we have

`E = ((6.626xx10^-34color(white)(l)"J"•cancel("s"))(299792458(cancel("m"))/(cancel("s"))))/(6.55 xx 10^-6cancel("m"))`

`= color(red)(3.03 xx 10^-20` `color(red)("J"`