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Calculate the change in energy of an atom that emits a photon of wavelength 2.21 meters. (Planck’s constant is 6.626 x 10-34 joule seconds, the speed of light is 2.998 x 108 m/s)

Respuesta :

proxy4
  Energy of photon = h x f 
f = v/wavelength 
f = 3x10^8 / 2.21 = 1.357x10^8 
E = 8.99x10^-26 joules i hope i helped u
laurahai

Answer:

The atom's energy will decrease by 8.99 x 10⁻²⁶ J.

Explanation:

The atom's energy will go down by an amount equal to the energy of the photon that it emits.

The energy E of a photon is inversely proportional to its wavelength according to the following equation, where h is Planck's constant and c is the speed of light:

E = (hc)/λ

We substitute the values into the equation to solve for the energy:

E = (hc)/λ = (6.626 x 10⁻³⁴ J·s)(2.998 x 10⁸ m·s⁻¹) / 2.21 m

E = 8.99 x 10 ⁻²⁶ J

Thus, the atom's energy will decrease by 8.99 x 10⁻²⁶ J.

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