Question

What wavelength (in nm) of photon absorbed when an electron transitions from then = 2 to the n = 21 state?Using-2.998 for light

Answer 1

ANSWER

The wavelength of the photon is 367.87 nm

EXPLANATION

Given information

$Speed\text{ of light = 2.998 }\times\text{ 10}^8\text{ m/s}$

To find the wavelength of the photon absorbed, we will need to find the change in energy in transition using the below formula

$E_n\text{ = -2.18 }\times\text{ 10}^{-18}\text{ }(\frac{1}{n^2})$

From the question, the energy states given are from n = 21 to n =2

So, we can calculate the value of energy below

$E_{21}\text{ - E}_2\text{ = }\Delta E$$\begin{gathered} \Delta E\text{ = -2.18 }\times\text{ 10}^{-18}\text{ }(\frac{1}{21^2}-\frac{1}{2^2}) \\ \Delta E\text{ = -2.18 }\times10^{-18}(\frac{1}{441}-\frac{1}{4}) \\ \Delta E\text{ = -2.18}\times10^{-18}(0.0022675737\text{ - 0.25}) \\ \Delta E\text{ = -2.18}\times10^{-18}\text{ }(-0.2477324263) \\ \Delta E\text{ = -2.18}\times10^{-18}\times(-2.477324263\times10^{-1}) \\ \Delta E\text{ = 2.18}\times2.477324263\times10^{-18-1} \\ \Delta E\text{ = 5.400}\times10^{-19}J \end{gathered}$

Since we have gotten the value of the energy, hence, we can now calculate the wavelength of the photon using the below formula

$\begin{gathered} \Delta E=\frac{hc}{\lambda} \\ Where \\ h\text{ = Planck's constant} \\ c\text{ = speed of light} \\ \lambda\text{ = wavelength} \end{gathered}$

Recall,

h = 6.626 x 10^-34 J.s

c = 2.998 x 10^8 m/s

$\begin{gathered} \lambda=\frac{hc}{\Delta E} \\ \frac{}{} \\ \lambda=\frac{6.626\times10^{-34}\times2.998\times10^8}{5.40\times10^{-19}} \\ \lambda=\frac{6.626\times2.998\times10^{-34+8}}{5.40\times10^{-19}} \\ \\ \lambda=\frac{19.865\times10^{-26+19}}{5.400} \\ \lambda=\frac{19.865}{5.40}\times10^{-26+19} \\ \lambda=3.6787\text{ }\times10^{-7} \\ \lambda=\text{ 367.87nm} \end{gathered}$

Hence, the wavelength of the photon is 367.87 nm