Question

What is the change in energy of an atom if the wavelength of the photon absorbed by the atom is equal to 4.64 x 10-7 meters? (Planck’s constant is 6.626 x 10-34 joule seconds; the speed of light is 2.998 x 108 m/s)?

Answer 1

Answer : The change in energy of an atom is, $4.28\times 10^{-19}J$

Solution : Given,

Wavelength of the photon = $4.64\times 10^{-7}m$

Formula used :

$E=h\times \frac{c}{\lambda}$

where,

E = energy of an atom

h = Planck’s constant = $6.626\times 10^{-34}Js$

c = speed of light = $2.998\times 10^{8}m/s$

$\lambda$ = wavelength of photon

Now put all the given values in the above formula, we get the energy of an atom.

$E=(6.626\times 10^{-34}Js)\times \frac{2.998\times 10^8m/s}{4.64\times 10^{-7}m}$

$E=4.28\times 10^{-19}J$

Therefore, the change in energy of an atom is, $4.28\times 10^{-19}J$

Answer 2

Hello!

We have the following data:

v (speed of light) = $2.998*10^8\:m/s$

λ (wavelength) = $4.64*10^{-7}\:m$

First, let's find the frequency of the wave, let's see:

$f = \dfrac{v}{\lambda}$

$f = \dfrac{2.998*10^8\:\diagup\!\!\!\!\!m/s}{4.64*10^{-7}\:\diagup\!\!\!\!\!m}$

$f \approx 0.6461*10^{8-(-7)}$

$f \approx 0.6461*10^{8+7}$

$f \approx 0.6461*10^{15}$

$\boxed{f \approx 6.461*10^{14}\:Hz}$

Now, find the Energy of the Photon by Planck's Equation, given:

E (photon energy) =? (in Joule)

h (Planck's constant) = $6.626*10^{-34}\:J * s$

f (radiation frequency) = $6.461*10^{14}\:Hz$

Therefore, we have:

$E = h*f$

$E = 6.626*10^{-34}*6.461*10^{14}$

$E \approx 42.81*10^{-34+14}$

$E \approx 42.81*10^{-20}$

$\boxed{\boxed{E \approx 4.281*10^{-19}\:Joule}}\end{array}}\qquad\checkmark$

I Hope this helps, greetings ... DexteR! =)