A local FM radio station broadcasts at an energy of 6.04E-29 kJ/photon. Calculate the frequency at which it is broadcasting.

9.11×10⁷ Hz

Explanation:

From the question given above the following data were obtained:

Energy (E) = 6.04×10¯²⁹ KJ

Frequency (f) =?

Next, we shall convert 6.04×10¯²⁹ KJ to joule (J). This can be obtained as follow:

1 KJ = 1000 J

Therefore,

6.04×10¯²⁹ KJ = 6.04×10¯²⁹ KJ × 1000 J / 1 KJ

6.04×10¯²⁹ KJ = 6.04×10¯²⁶ J

Thus, 6.04×10¯²⁹ KJ is equivalent to 6.04×10¯²⁶ J.

Finally, we shall determine the frequency at which the FM radio station is broadcasting as follow:

Energy (E) = 6.04×10¯²⁶ J

Planck's constant (h) = 6.63×10¯³⁴ Js

Frequency (f) =?

E = hf

6.04×10¯²⁶ = 6.63×10¯³⁴ × f

Divide both side by 6.63×10¯³⁴

f = 6.04×10¯²⁶ / 6.63×10¯³⁴

f = 9.11×10⁷ Hz

Therefore, the frequency at which the FM radio station is broadcasting is 9.11×10⁷ Hz

your answer is a. the enthalpy of formation for a pure element in its standard state is always positive.