How do you calculate the wavelength of a photon that has an energy of $1.257 * 10^(-24)$ $"J"$ ?

The inverse relationship that exists between wavelength and frequency is described by the equation $color(blue)( lamda * nu = c)" "$, where $lamda$ - the frequency of...

First find the frequency of the photon from the equation $c=flambda$ $therefore f=c/lambda=(3xx10^8m//s)/(0.050xx10^(-2)m)=6xx10^11Hz$. Energy of such photons is quantized and given by $E=hf$ $=(6.6xx10^(-34))(6xx10^11)$ $=3.96xx10^(-22)J$

Let us calculate the frequency $nu$ of the photon. We have the expression between the velocity of light in vacuum $c=nuxxlambda$ or $nu=c/lambda$......(1) where $lambda$ is the wave-length of...

Let $c = "the speed of light" = 3.0 xx 10^8 m/s$ Let $lambda = "the wavelength" = 4.257 xx 10^-9 m$ Let $f = $ the unknown frequency (in...

To find the energy of a photon you use the Planck Expression: $sf(E=hf)$ $sf(h)$ is the Planck Constant which = $sf(6.63xx10^(-34)color(white)(x)J.s)$ $sf(f)$ is the frequency This becomes: $sf(E=(hc)/lambda)$ $sf(c)$ is...

Wavelength You use the Rydberg formula to calculate the wavelength, λ: $color(blue)(bar(ul(|color(white)(a/a) 1/λ = RZ^2(1/n_2^2 -1/n_1^2)color(white)(a/a)|)))" "$ where $R =$ the Rydberg constant ($1.097 × 10^7color(white)(l)...

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...

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)...