Microwave ovens emit microwave energy with a wavelength of 12.2 cm. What is the energy of exactly one photon of this microwave radiation?

For electromagnetic radiation: $c=flambda$ where $c$ is the speed of light, $f$ is the frequency, $lambda$ is the wavelength. Rearranging: $lambda = c/f=(3.0xx10^8)/(3.0xx10^10)=10^-2$ $m$ The energy of a photon is...

The energy of a photon is actually directly proportional to its frequency as given by the Planck - Einstein relation $color(blue)(ul(color(black)(E = h * nu)))$ Here...

All you have to do here is to use the Planck - Einstein relation, which as you know establishes a relationship between the frequency of a photon and the energy...

$E= h xx f$ $E$ is energy in joules $h$is $ = 6.626 xx 10^-34("m"^2"kg")/"s"$ $f$ is frequency, in $"Hz"$ or $"s"^(-1)$ So $E= 6.626 xx 10^-34 ("m"^2...

The energy of a photon is $E=hf$ The energy is $E$ The frequency is $f$ is $h=6.626*10^-34Js$ But $f=c/lambda$ The speed of light is $c=3*10^8ms^-1$ The wavelength is...

Phosphorus has $15$. As such its electronic configuration in ground state is $1s^2 2s^2 2p^6 3s^2 3p^3$ The following electronic configurations could be excited states $1s^2 2s^2...

We're asked to calculate the energy of one photon of a green light, given its wavelength of $5200$ $"Å"$. To do this, we can use the equation...

Relativistic effects are present here... but I assume that the transition from level $a$ to level $g$: has $DeltaL = pm 1$ (the total change in angular momentum is...

About $"9.78 keV"$, assuming $100%$ transfer of energy. The de Broglie wavelength $lambda$ is given by $lambda = h/p = h/(mv)$, where: $h = 6.626 xx 10^(-34)...

This is a very difficult problem. What makes it difficult is the number of steps involved in getting to the answer. The outline of how we are going to solve...